Practice · Match coefficients of like terms · Sensim Math

Variant 1 easy

The equation $3x(Bx+5)+A(Bx+5)=6x^2+Cx-10$ holds for every value of $x$ . Find the value of $A+B+C$ . ( $A$ , $B$ , and $C$ are constants.)

Your answer

Show solution

1 · Understandwhat's really being asked

Two expressions in $x$ are equal for every $x$ . One side has the unknown constants $A$ and $B$ inside it; the other has $C$ . Find $A+B+C$ .

Givens

$3x(Bx+5)+A(Bx+5)=6x^2+Cx-10$ for every $x$ .
$A$ , $B$ , $C$ are constants.

Unknowns

The value of $A+B+C$ .

Constraints

Two polynomials that are equal for every $x$ have the same coefficient on each power of $x$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#15 Organize Information in More Ways

Expand the left side, sort it by powers of $x$ , and set each coefficient equal to the matching one on the right. That turns one identity into three small equations.

3 · Execute5 carry out the plan

1Expand the left side

#7 Identify Subproblems A-APR.A.1

Distribute

3x

over

(Bx+5)

and

A

over

(Bx+5)

.

3x(Bx+5)+A(Bx+5)=3Bx^2+15x+ABx+5A

Just the distributive property, twice.

2Sort by powers of x

#15 Organize Information in More Ways A-APR.A.1

Collect the two

x

terms together;

x \times x = x^2

gives the square term.

3Bx^2+(15+AB)x+5A

Now each power of

x

appears once.

3Match the coefficients

#13 Convert to Algebra 6.EE.A.4

Equal for every

x

means equal coefficients:

x^2

with

x^2

,

x

with

x

, constant with constant.

3B=6,\ 15+AB=C,\ 5A=-10

Three equations come out of one.

4Solve the small equations

#13 Convert to Algebra 6.EE.A.4

The first and last give

B

and

A

at once; then the middle one gives

C

.

B=2,\ A=-2,\ C=15+(-2)\times 2=11

Solve the easy ones first, then use them.

5Add them

#7 Identify Subproblems A-APR.A.1

Add the three constants.

A+B+C=-2+2+11=11

A

and

B

cancel each other.

Answer:

11

4 · Reviewdoes it hold up?

Put the values back: $3x(2x+5)-2(2x+5)=6x^2+15x-4x-10=6x^2+11x-10$ , which matches the right side with $C=11$ .

Another way: Plug in convenient values of $x$ . With $x=0$ : $5A=-10$ , so $A=-2$ . The left side is $(3x+A)(Bx+5)$ , so its $x^2$ coefficient $3B$ must be $6$ , giving $B=2$ ; then $x=1$ gives $(3-2)(2+5)=7=6+C-10$ , so $C=11$ .

Standardsmin grade 9

A-APR.A.1 Polynomials are closed under addition, subtraction, multiplication — Multiplying out both products and collecting the polynomial by powers of $x$.
6.EE.A.4 Identify equivalent expressions — Equal expressions have equal coefficients on like terms.

💡Takeaway. If two polynomials are equal for every

x

, line up their terms and the unknowns fall out one by one.

Variant 2 easy

The equation $5x(Bx+3)+A(Bx+3)=10x^2+Cx+6$ holds for every value of $x$ . Find the value of $A+B+C$ . ( $A$ , $B$ , and $C$ are constants.)

Your answer

Show solution

1 · Understandwhat's really being asked

Two expressions in $x$ are equal for every $x$ . One side has the unknown constants $A$ and $B$ inside it; the other has $C$ . Find $A+B+C$ .

Givens

$5x(Bx+3)+A(Bx+3)=10x^2+Cx+6$ for every $x$ .
$A$ , $B$ , $C$ are constants.

Unknowns

The value of $A+B+C$ .

Constraints

Two polynomials that are equal for every $x$ have the same coefficient on each power of $x$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#15 Organize Information in More Ways

Expand the left side, sort it by powers of $x$ , and set each coefficient equal to the matching one on the right. That turns one identity into three small equations.

3 · Execute5 carry out the plan

1Expand the left side

#7 Identify Subproblems A-APR.A.1

Distribute

5x

over

(Bx+3)

and

A

over

(Bx+3)

.

5x(Bx+3)+A(Bx+3)=5Bx^2+15x+ABx+3A

Just the distributive property, twice.

2Sort by powers of x

#15 Organize Information in More Ways A-APR.A.1

Collect the two

x

terms together;

x \times x = x^2

gives the square term.

5Bx^2+(15+AB)x+3A

Now each power of

x

appears once.

3Match the coefficients

#13 Convert to Algebra 6.EE.A.4

Equal for every

x

means equal coefficients:

x^2

with

x^2

,

x

with

x

, constant with constant.

5B=10,\ 15+AB=C,\ 3A=6

Three equations come out of one.

4Solve the small equations

#13 Convert to Algebra 6.EE.A.4

The first and last give

B

and

A

at once; then the middle one gives

C

.

B=2,\ A=2,\ C=15+2\times 2=19

Solve the easy ones first, then use them.

5Add them

#7 Identify Subproblems A-APR.A.1

Add the three constants.

A+B+C=2+2+19=23

A

and

B

together give

4

.

Answer:

23

4 · Reviewdoes it hold up?

Put the values back: $5x(2x+3)+2(2x+3)=10x^2+15x+4x+6=10x^2+19x+6$ , which matches the right side with $C=19$ .

Another way: Plug in convenient values of $x$ . With $x=0$ : $3A=6$ , so $A=2$ . The left side is $(5x+A)(Bx+3)$ , so its $x^2$ coefficient $5B$ must be $10$ , giving $B=2$ ; then $x=1$ gives $(5+2)(2+3)=35=10+C+6$ , so $C=19$ .

Standardsmin grade 9

A-APR.A.1 Polynomials are closed under addition, subtraction, multiplication — Multiplying out both products and collecting the polynomial by powers of $x$.
6.EE.A.4 Identify equivalent expressions — Equal expressions have equal coefficients on like terms.

💡Takeaway. If two polynomials are equal for every

x

, line up their terms and the unknowns fall out one by one.

Variant 3 easy

The equation $4x(Bx+5)+A(Bx+5)=12x^2+Cx+5$ holds for every value of $x$ . Find the value of $A+B+C$ . ( $A$ , $B$ , and $C$ are constants.)

Your answer

Show solution

1 · Understandwhat's really being asked

Two expressions in $x$ are equal for every $x$ . One side has the unknown constants $A$ and $B$ inside it; the other has $C$ . Find $A+B+C$ .

Givens

$4x(Bx+5)+A(Bx+5)=12x^2+Cx+5$ for every $x$ .
$A$ , $B$ , $C$ are constants.

Unknowns

The value of $A+B+C$ .

Constraints

Two polynomials that are equal for every $x$ have the same coefficient on each power of $x$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#15 Organize Information in More Ways

Expand the left side, sort it by powers of $x$ , and set each coefficient equal to the matching one on the right. That turns one identity into three small equations.

3 · Execute5 carry out the plan

1Expand the left side

#7 Identify Subproblems A-APR.A.1

Distribute

4x

over

(Bx+5)

and

A

over

(Bx+5)

.

4x(Bx+5)+A(Bx+5)=4Bx^2+20x+ABx+5A

Just the distributive property, twice.

2Sort by powers of x

#15 Organize Information in More Ways A-APR.A.1

Collect the two

x

terms together;

x \times x = x^2

gives the square term.

4Bx^2+(20+AB)x+5A

Now each power of

x

appears once.

3Match the coefficients

#13 Convert to Algebra 6.EE.A.4

Equal for every

x

means equal coefficients:

x^2

with

x^2

,

x

with

x

, constant with constant.

4B=12,\ 20+AB=C,\ 5A=5

Three equations come out of one.

4Solve the small equations

#13 Convert to Algebra 6.EE.A.4

The first and last give

B

and

A

at once; then the middle one gives

C

.

B=3,\ A=1,\ C=20+1\times 3=23

Solve the easy ones first, then use them.

5Add them

#7 Identify Subproblems A-APR.A.1

Add the three constants.

A+B+C=1+3+23=27

A

and

B

together give

4

.

Answer:

27

4 · Reviewdoes it hold up?

Put the values back: $4x(3x+5)+1(3x+5)=12x^2+20x+3x+5=12x^2+23x+5$ , which matches the right side with $C=23$ .

Another way: Plug in convenient values of $x$ . With $x=0$ : $5A=5$ , so $A=1$ . The left side is $(4x+A)(Bx+5)$ , so its $x^2$ coefficient $4B$ must be $12$ , giving $B=3$ ; then $x=1$ gives $(4+1)(3+5)=40=12+C+5$ , so $C=23$ .

Standardsmin grade 9

A-APR.A.1 Polynomials are closed under addition, subtraction, multiplication — Multiplying out both products and collecting the polynomial by powers of $x$.
6.EE.A.4 Identify equivalent expressions — Equal expressions have equal coefficients on like terms.

💡Takeaway. If two polynomials are equal for every

x

, line up their terms and the unknowns fall out one by one.

Variant 4 easy

The equation $2x(Bx+3)+A(Bx+3)=6x^2+Cx+12$ holds for every value of $x$ . Find the value of $A+B+C$ . ( $A$ , $B$ , and $C$ are constants.)

Your answer

Show solution

1 · Understandwhat's really being asked

Two expressions in $x$ are equal for every $x$ . One side has the unknown constants $A$ and $B$ inside it; the other has $C$ . Find $A+B+C$ .

Givens

$2x(Bx+3)+A(Bx+3)=6x^2+Cx+12$ for every $x$ .
$A$ , $B$ , $C$ are constants.

Unknowns

The value of $A+B+C$ .

Constraints

Two polynomials that are equal for every $x$ have the same coefficient on each power of $x$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#15 Organize Information in More Ways

Expand the left side, sort it by powers of $x$ , and set each coefficient equal to the matching one on the right. That turns one identity into three small equations.

3 · Execute5 carry out the plan

1Expand the left side

#7 Identify Subproblems A-APR.A.1

Distribute

2x

over

(Bx+3)

and

A

over

(Bx+3)

.

2x(Bx+3)+A(Bx+3)=2Bx^2+6x+ABx+3A

Just the distributive property, twice.

2Sort by powers of x

#15 Organize Information in More Ways A-APR.A.1

Collect the two

x

terms together;

x \times x = x^2

gives the square term.

2Bx^2+(6+AB)x+3A

Now each power of

x

appears once.

3Match the coefficients

#13 Convert to Algebra 6.EE.A.4

Equal for every

x

means equal coefficients:

x^2

with

x^2

,

x

with

x

, constant with constant.

2B=6,\ 6+AB=C,\ 3A=12

Three equations come out of one.

4Solve the small equations

#13 Convert to Algebra 6.EE.A.4

The first and last give

B

and

A

at once; then the middle one gives

C

.

B=3,\ A=4,\ C=6+4\times 3=18

Solve the easy ones first, then use them.

5Add them

#7 Identify Subproblems A-APR.A.1

Add the three constants.

A+B+C=4+3+18=25

A

and

B

together give

7

.

Answer:

25

4 · Reviewdoes it hold up?

Put the values back: $2x(3x+3)+4(3x+3)=6x^2+6x+12x+12=6x^2+18x+12$ , which matches the right side with $C=18$ .

Another way: Plug in convenient values of $x$ . With $x=0$ : $3A=12$ , so $A=4$ . The left side is $(2x+A)(Bx+3)$ , so its $x^2$ coefficient $2B$ must be $6$ , giving $B=3$ ; then $x=1$ gives $(2+4)(3+3)=36=6+C+12$ , so $C=18$ .

Standardsmin grade 9

A-APR.A.1 Polynomials are closed under addition, subtraction, multiplication — Multiplying out both products and collecting the polynomial by powers of $x$.
6.EE.A.4 Identify equivalent expressions — Equal expressions have equal coefficients on like terms.

💡Takeaway. If two polynomials are equal for every

x

, line up their terms and the unknowns fall out one by one.

Variant 5 medium

The equation $3x(Bx+2)+A(Bx+2)=12x^2+Cx-14$ holds for every value of $x$ . Find the value of $A+B+C$ . ( $A$ , $B$ , and $C$ are constants.)

Your answer

Show solution

1 · Understandwhat's really being asked

Two expressions in $x$ are equal for every $x$ . One side has the unknown constants $A$ and $B$ inside it; the other has $C$ . Find $A+B+C$ .

Givens

$3x(Bx+2)+A(Bx+2)=12x^2+Cx-14$ for every $x$ .
$A$ , $B$ , $C$ are constants.

Unknowns

The value of $A+B+C$ .

Constraints

Two polynomials that are equal for every $x$ have the same coefficient on each power of $x$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#15 Organize Information in More Ways

Expand the left side, sort it by powers of $x$ , and set each coefficient equal to the matching one on the right. That turns one identity into three small equations.

3 · Execute5 carry out the plan

1Expand the left side

#7 Identify Subproblems A-APR.A.1

Distribute

3x

over

(Bx+2)

and

A

over

(Bx+2)

.

3x(Bx+2)+A(Bx+2)=3Bx^2+6x+ABx+2A

Just the distributive property, twice.

2Sort by powers of x

#15 Organize Information in More Ways A-APR.A.1

Collect the two

x

terms together;

x \times x = x^2

gives the square term.

3Bx^2+(6+AB)x+2A

Now each power of

x

appears once.

3Match the coefficients

#13 Convert to Algebra 6.EE.A.4

Equal for every

x

means equal coefficients:

x^2

with

x^2

,

x

with

x

, constant with constant.

3B=12,\ 6+AB=C,\ 2A=-14

Three equations come out of one.

4Solve the small equations

#13 Convert to Algebra 6.EE.A.4

The first and last give

B

and

A

at once; then the middle one gives

C

.

B=4,\ A=-7,\ C=6+(-7)\times 4=-22

Solve the easy ones first, then use them.

5Add them

#7 Identify Subproblems A-APR.A.1

Add the three constants.

A+B+C=-7+4-22=-25

A

and

B

together give

-3

.

Answer:

-25

4 · Reviewdoes it hold up?

Put the values back: $3x(4x+2)-7(4x+2)=12x^2+6x-28x-14=12x^2-22x-14$ , which matches the right side with $C=-22$ .

Another way: Plug in convenient values of $x$ . With $x=0$ : $2A=-14$ , so $A=-7$ . The left side is $(3x+A)(Bx+2)$ , so its $x^2$ coefficient $3B$ must be $12$ , giving $B=4$ ; then $x=1$ gives $(3-7)(4+2)=-24=12+C-14$ , so $C=-22$ .

Standardsmin grade 9

A-APR.A.1 Polynomials are closed under addition, subtraction, multiplication — Multiplying out both products and collecting the polynomial by powers of $x$.
6.EE.A.4 Identify equivalent expressions — Equal expressions have equal coefficients on like terms.

💡Takeaway. If two polynomials are equal for every

x

, line up their terms and the unknowns fall out one by one.

Variant 6 medium

The equation $2x(Bx+5)+A(Bx+5)=8x^2+Cx-15$ holds for every value of $x$ . Find the value of $A+B+C$ . ( $A$ , $B$ , and $C$ are constants.)

Your answer

Show solution

1 · Understandwhat's really being asked

Two expressions in $x$ are equal for every $x$ . One side has the unknown constants $A$ and $B$ inside it; the other has $C$ . Find $A+B+C$ .

Givens

$2x(Bx+5)+A(Bx+5)=8x^2+Cx-15$ for every $x$ .
$A$ , $B$ , $C$ are constants.

Unknowns

The value of $A+B+C$ .

Constraints

Two polynomials that are equal for every $x$ have the same coefficient on each power of $x$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#15 Organize Information in More Ways

Expand the left side, sort it by powers of $x$ , and set each coefficient equal to the matching one on the right. That turns one identity into three small equations.

3 · Execute5 carry out the plan

1Expand the left side

#7 Identify Subproblems A-APR.A.1

Distribute

2x

over

(Bx+5)

and

A

over

(Bx+5)

.

2x(Bx+5)+A(Bx+5)=2Bx^2+10x+ABx+5A

Just the distributive property, twice.

2Sort by powers of x

#15 Organize Information in More Ways A-APR.A.1

Collect the two

x

terms together;

x \times x = x^2

gives the square term.

2Bx^2+(10+AB)x+5A

Now each power of

x

appears once.

3Match the coefficients

#13 Convert to Algebra 6.EE.A.4

Equal for every

x

means equal coefficients:

x^2

with

x^2

,

x

with

x

, constant with constant.

2B=8,\ 10+AB=C,\ 5A=-15

Three equations come out of one.

4Solve the small equations

#13 Convert to Algebra 6.EE.A.4

The first and last give

B

and

A

at once; then the middle one gives

C

.

B=4,\ A=-3,\ C=10+(-3)\times 4=-2

Solve the easy ones first, then use them.

5Add them

#7 Identify Subproblems A-APR.A.1

Add the three constants.

A+B+C=-3+4-2=-1

A

and

B

together give

1

.

Answer:

-1

4 · Reviewdoes it hold up?

Put the values back: $2x(4x+5)-3(4x+5)=8x^2+10x-12x-15=8x^2-2x-15$ , which matches the right side with $C=-2$ .

Another way: Plug in convenient values of $x$ . With $x=0$ : $5A=-15$ , so $A=-3$ . The left side is $(2x+A)(Bx+5)$ , so its $x^2$ coefficient $2B$ must be $8$ , giving $B=4$ ; then $x=1$ gives $(2-3)(4+5)=-9=8+C-15$ , so $C=-2$ .

Standardsmin grade 9

A-APR.A.1 Polynomials are closed under addition, subtraction, multiplication — Multiplying out both products and collecting the polynomial by powers of $x$.
6.EE.A.4 Identify equivalent expressions — Equal expressions have equal coefficients on like terms.

💡Takeaway. If two polynomials are equal for every

x

, line up their terms and the unknowns fall out one by one.

Variant 7 medium

The equation $3x(Bx+4)+A(Bx+4)=15x^2+Cx+8$ holds for every value of $x$ . Find the value of $A+B+C$ . ( $A$ , $B$ , and $C$ are constants.)

Your answer

Show solution

1 · Understandwhat's really being asked

Two expressions in $x$ are equal for every $x$ . One side has the unknown constants $A$ and $B$ inside it; the other has $C$ . Find $A+B+C$ .

Givens

$3x(Bx+4)+A(Bx+4)=15x^2+Cx+8$ for every $x$ .
$A$ , $B$ , $C$ are constants.

Unknowns

The value of $A+B+C$ .

Constraints

Two polynomials that are equal for every $x$ have the same coefficient on each power of $x$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#15 Organize Information in More Ways

Expand the left side, sort it by powers of $x$ , and set each coefficient equal to the matching one on the right. That turns one identity into three small equations.

3 · Execute5 carry out the plan

1Expand the left side

#7 Identify Subproblems A-APR.A.1

Distribute

3x

over

(Bx+4)

and

A

over

(Bx+4)

.

3x(Bx+4)+A(Bx+4)=3Bx^2+12x+ABx+4A

Just the distributive property, twice.

2Sort by powers of x

#15 Organize Information in More Ways A-APR.A.1

Collect the two

x

terms together;

x \times x = x^2

gives the square term.

3Bx^2+(12+AB)x+4A

Now each power of

x

appears once.

3Match the coefficients

#13 Convert to Algebra 6.EE.A.4

Equal for every

x

means equal coefficients:

x^2

with

x^2

,

x

with

x

, constant with constant.

3B=15,\ 12+AB=C,\ 4A=8

Three equations come out of one.

4Solve the small equations

#13 Convert to Algebra 6.EE.A.4

The first and last give

B

and

A

at once; then the middle one gives

C

.

B=5,\ A=2,\ C=12+2\times 5=22

Solve the easy ones first, then use them.

5Add them

#7 Identify Subproblems A-APR.A.1

Add the three constants.

A+B+C=2+5+22=29

A

and

B

together give

7

.

Answer:

29

4 · Reviewdoes it hold up?

Put the values back: $3x(5x+4)+2(5x+4)=15x^2+12x+10x+8=15x^2+22x+8$ , which matches the right side with $C=22$ .

Another way: Plug in convenient values of $x$ . With $x=0$ : $4A=8$ , so $A=2$ . The left side is $(3x+A)(Bx+4)$ , so its $x^2$ coefficient $3B$ must be $15$ , giving $B=5$ ; then $x=1$ gives $(3+2)(5+4)=45=15+C+8$ , so $C=22$ .

Standardsmin grade 9

A-APR.A.1 Polynomials are closed under addition, subtraction, multiplication — Multiplying out both products and collecting the polynomial by powers of $x$.
6.EE.A.4 Identify equivalent expressions — Equal expressions have equal coefficients on like terms.

💡Takeaway. If two polynomials are equal for every

x

, line up their terms and the unknowns fall out one by one.

Variant 8 hard

The equation $5x(Bx+2)+A(Bx+2)=15x^2+Cx-8$ holds for every value of $x$ . Find the value of $A+B+C$ . ( $A$ , $B$ , and $C$ are constants.)

Your answer

Show solution

1 · Understandwhat's really being asked

Two expressions in $x$ are equal for every $x$ . One side has the unknown constants $A$ and $B$ inside it; the other has $C$ . Find $A+B+C$ .

Givens

$5x(Bx+2)+A(Bx+2)=15x^2+Cx-8$ for every $x$ .
$A$ , $B$ , $C$ are constants.

Unknowns

The value of $A+B+C$ .

Constraints

Two polynomials that are equal for every $x$ have the same coefficient on each power of $x$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#15 Organize Information in More Ways

Expand the left side, sort it by powers of $x$ , and set each coefficient equal to the matching one on the right. That turns one identity into three small equations.

3 · Execute5 carry out the plan

1Expand the left side

#7 Identify Subproblems A-APR.A.1

Distribute

5x

over

(Bx+2)

and

A

over

(Bx+2)

.

5x(Bx+2)+A(Bx+2)=5Bx^2+10x+ABx+2A

Just the distributive property, twice.

2Sort by powers of x

#15 Organize Information in More Ways A-APR.A.1

Collect the two

x

terms together;

x \times x = x^2

gives the square term.

5Bx^2+(10+AB)x+2A

Now each power of

x

appears once.

3Match the coefficients

#13 Convert to Algebra 6.EE.A.4

Equal for every

x

means equal coefficients:

x^2

with

x^2

,

x

with

x

, constant with constant.

5B=15,\ 10+AB=C,\ 2A=-8

Three equations come out of one.

4Solve the small equations

#13 Convert to Algebra 6.EE.A.4

The first and last give

B

and

A

at once; then the middle one gives

C

.

B=3,\ A=-4,\ C=10+(-4)\times 3=-2

Solve the easy ones first, then use them.

5Add them

#7 Identify Subproblems A-APR.A.1

Add the three constants.

A+B+C=-4+3-2=-3

A

and

B

together give

-1

.

Answer:

-3

4 · Reviewdoes it hold up?

Put the values back: $5x(3x+2)-4(3x+2)=15x^2+10x-12x-8=15x^2-2x-8$ , which matches the right side with $C=-2$ .

Another way: Plug in convenient values of $x$ . With $x=0$ : $2A=-8$ , so $A=-4$ . The left side is $(5x+A)(Bx+2)$ , so its $x^2$ coefficient $5B$ must be $15$ , giving $B=3$ ; then $x=1$ gives $(5-4)(3+2)=5=15+C-8$ , so $C=-2$ .

Standardsmin grade 9

A-APR.A.1 Polynomials are closed under addition, subtraction, multiplication — Multiplying out both products and collecting the polynomial by powers of $x$.
6.EE.A.4 Identify equivalent expressions — Equal expressions have equal coefficients on like terms.

💡Takeaway. If two polynomials are equal for every

x

, line up their terms and the unknowns fall out one by one.

Variant 9 medium

The equation $4x(Bx+3)+A(Bx+3)=8x^2+Cx-15$ holds for every value of $x$ . Find the value of $A+B+C$ . ( $A$ , $B$ , and $C$ are constants.)

Your answer

Show solution

1 · Understandwhat's really being asked

Two expressions in $x$ are equal for every $x$ . One side has the unknown constants $A$ and $B$ inside it; the other has $C$ . Find $A+B+C$ .

Givens

$4x(Bx+3)+A(Bx+3)=8x^2+Cx-15$ for every $x$ .
$A$ , $B$ , $C$ are constants.

Unknowns

The value of $A+B+C$ .

Constraints

Two polynomials that are equal for every $x$ have the same coefficient on each power of $x$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#15 Organize Information in More Ways

Expand the left side, sort it by powers of $x$ , and set each coefficient equal to the matching one on the right. That turns one identity into three small equations.

3 · Execute5 carry out the plan

1Expand the left side

#7 Identify Subproblems A-APR.A.1

Distribute

4x

over

(Bx+3)

and

A

over

(Bx+3)

.

4x(Bx+3)+A(Bx+3)=4Bx^2+12x+ABx+3A

Just the distributive property, twice.

2Sort by powers of x

#15 Organize Information in More Ways A-APR.A.1

Collect the two

x

terms together;

x \times x = x^2

gives the square term.

4Bx^2+(12+AB)x+3A

Now each power of

x

appears once.

3Match the coefficients

#13 Convert to Algebra 6.EE.A.4

Equal for every

x

means equal coefficients:

x^2

with

x^2

,

x

with

x

, constant with constant.

4B=8,\ 12+AB=C,\ 3A=-15

Three equations come out of one.

4Solve the small equations

#13 Convert to Algebra 6.EE.A.4

The first and last give

B

and

A

at once; then the middle one gives

C

.

B=2,\ A=-5,\ C=12+(-5)\times 2=2

Solve the easy ones first, then use them.

5Add them

#7 Identify Subproblems A-APR.A.1

Add the three constants.

A+B+C=-5+2+2=-1

A

and

B

together give

-3

.

Answer:

-1

4 · Reviewdoes it hold up?

Put the values back: $4x(2x+3)-5(2x+3)=8x^2+12x-10x-15=8x^2+2x-15$ , which matches the right side with $C=2$ .

Another way: Plug in convenient values of $x$ . With $x=0$ : $3A=-15$ , so $A=-5$ . The left side is $(4x+A)(Bx+3)$ , so its $x^2$ coefficient $4B$ must be $8$ , giving $B=2$ ; then $x=1$ gives $(4-5)(2+3)=-5=8+C-15$ , so $C=2$ .

Standardsmin grade 9

A-APR.A.1 Polynomials are closed under addition, subtraction, multiplication — Multiplying out both products and collecting the polynomial by powers of $x$.
6.EE.A.4 Identify equivalent expressions — Equal expressions have equal coefficients on like terms.

💡Takeaway. If two polynomials are equal for every

x

, line up their terms and the unknowns fall out one by one.

Variant 10 hard

The equation $3x(Bx+6)+A(Bx+6)=18x^2+Cx-6$ holds for every value of $x$ . Find the value of $A+B+C$ . ( $A$ , $B$ , and $C$ are constants.)

Your answer

Show solution

1 · Understandwhat's really being asked

Two expressions in $x$ are equal for every $x$ . One side has the unknown constants $A$ and $B$ inside it; the other has $C$ . Find $A+B+C$ .

Givens

$3x(Bx+6)+A(Bx+6)=18x^2+Cx-6$ for every $x$ .
$A$ , $B$ , $C$ are constants.

Unknowns

The value of $A+B+C$ .

Constraints

Two polynomials that are equal for every $x$ have the same coefficient on each power of $x$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#15 Organize Information in More Ways

Expand the left side, sort it by powers of $x$ , and set each coefficient equal to the matching one on the right. That turns one identity into three small equations.

3 · Execute5 carry out the plan

1Expand the left side

#7 Identify Subproblems A-APR.A.1

Distribute

3x

over

(Bx+6)

and

A

over

(Bx+6)

.

3x(Bx+6)+A(Bx+6)=3Bx^2+18x+ABx+6A

Just the distributive property, twice.

2Sort by powers of x

#15 Organize Information in More Ways A-APR.A.1

Collect the two

x

terms together;

x \times x = x^2

gives the square term.

3Bx^2+(18+AB)x+6A

Now each power of

x

appears once.

3Match the coefficients

#13 Convert to Algebra 6.EE.A.4

Equal for every

x

means equal coefficients:

x^2

with

x^2

,

x

with

x

, constant with constant.

3B=18,\ 18+AB=C,\ 6A=-6

Three equations come out of one.

4Solve the small equations

#13 Convert to Algebra 6.EE.A.4

The first and last give

B

and

A

at once; then the middle one gives

C

.

B=6,\ A=-1,\ C=18+(-1)\times 6=12

Solve the easy ones first, then use them.

5Add them

#7 Identify Subproblems A-APR.A.1

Add the three constants.

A+B+C=-1+6+12=17

A

and

B

together give

5

.

Answer:

17

4 · Reviewdoes it hold up?

Put the values back: $3x(6x+6)-1(6x+6)=18x^2+18x-6x-6=18x^2+12x-6$ , which matches the right side with $C=12$ .

Another way: Plug in convenient values of $x$ . With $x=0$ : $6A=-6$ , so $A=-1$ . The left side is $(3x+A)(Bx+6)$ , so its $x^2$ coefficient $3B$ must be $18$ , giving $B=6$ ; then $x=1$ gives $(3-1)(6+6)=24=18+C-6$ , so $C=12$ .

Standardsmin grade 9

A-APR.A.1 Polynomials are closed under addition, subtraction, multiplication — Multiplying out both products and collecting the polynomial by powers of $x$.
6.EE.A.4 Identify equivalent expressions — Equal expressions have equal coefficients on like terms.

💡Takeaway. If two polynomials are equal for every

x

, line up their terms and the unknowns fall out one by one.

Variant 11 hard

The equation $2x(Bx+7)+A(Bx+7)=4x^2+Cx+21$ holds for every value of $x$ . Find the value of $A+B+C$ . ( $A$ , $B$ , and $C$ are constants.)

Your answer

Show solution

1 · Understandwhat's really being asked

Two expressions in $x$ are equal for every $x$ . One side has the unknown constants $A$ and $B$ inside it; the other has $C$ . Find $A+B+C$ .

Givens

$2x(Bx+7)+A(Bx+7)=4x^2+Cx+21$ for every $x$ .
$A$ , $B$ , $C$ are constants.

Unknowns

The value of $A+B+C$ .

Constraints

Two polynomials that are equal for every $x$ have the same coefficient on each power of $x$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#15 Organize Information in More Ways

Expand the left side, sort it by powers of $x$ , and set each coefficient equal to the matching one on the right. That turns one identity into three small equations.

3 · Execute5 carry out the plan

1Expand the left side

#7 Identify Subproblems A-APR.A.1

Distribute

2x

over

(Bx+7)

and

A

over

(Bx+7)

.

2x(Bx+7)+A(Bx+7)=2Bx^2+14x+ABx+7A

Just the distributive property, twice.

2Sort by powers of x

#15 Organize Information in More Ways A-APR.A.1

Collect the two

x

terms together;

x \times x = x^2

gives the square term.

2Bx^2+(14+AB)x+7A

Now each power of

x

appears once.

3Match the coefficients

#13 Convert to Algebra 6.EE.A.4

Equal for every

x

means equal coefficients:

x^2

with

x^2

,

x

with

x

, constant with constant.

2B=4,\ 14+AB=C,\ 7A=21

Three equations come out of one.

4Solve the small equations

#13 Convert to Algebra 6.EE.A.4

The first and last give

B

and

A

at once; then the middle one gives

C

.

B=2,\ A=3,\ C=14+3\times 2=20

Solve the easy ones first, then use them.

5Add them

#7 Identify Subproblems A-APR.A.1

Add the three constants.

A+B+C=3+2+20=25

A

and

B

together give

5

.

Answer:

25

4 · Reviewdoes it hold up?

Put the values back: $2x(2x+7)+3(2x+7)=4x^2+14x+6x+21=4x^2+20x+21$ , which matches the right side with $C=20$ .

Another way: Plug in convenient values of $x$ . With $x=0$ : $7A=21$ , so $A=3$ . The left side is $(2x+A)(Bx+7)$ , so its $x^2$ coefficient $2B$ must be $4$ , giving $B=2$ ; then $x=1$ gives $(2+3)(2+7)=45=4+C+21$ , so $C=20$ .

Standardsmin grade 9

A-APR.A.1 Polynomials are closed under addition, subtraction, multiplication — Multiplying out both products and collecting the polynomial by powers of $x$.
6.EE.A.4 Identify equivalent expressions — Equal expressions have equal coefficients on like terms.

💡Takeaway. If two polynomials are equal for every

x

, line up their terms and the unknowns fall out one by one.

Variant 12 hard

The equation $2x(Bx+4)+A(Bx+4)=10x^2+Cx-24$ holds for every value of $x$ . Find the value of $A+B+C$ . ( $A$ , $B$ , and $C$ are constants.)

Your answer

Show solution

1 · Understandwhat's really being asked

Two expressions in $x$ are equal for every $x$ . One side has the unknown constants $A$ and $B$ inside it; the other has $C$ . Find $A+B+C$ .

Givens

$2x(Bx+4)+A(Bx+4)=10x^2+Cx-24$ for every $x$ .
$A$ , $B$ , $C$ are constants.

Unknowns

The value of $A+B+C$ .

Constraints

Two polynomials that are equal for every $x$ have the same coefficient on each power of $x$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#15 Organize Information in More Ways

Expand the left side, sort it by powers of $x$ , and set each coefficient equal to the matching one on the right. That turns one identity into three small equations.

3 · Execute5 carry out the plan

1Expand the left side

#7 Identify Subproblems A-APR.A.1

Distribute

2x

over

(Bx+4)

and

A

over

(Bx+4)

.

2x(Bx+4)+A(Bx+4)=2Bx^2+8x+ABx+4A

Just the distributive property, twice.

2Sort by powers of x

#15 Organize Information in More Ways A-APR.A.1

Collect the two

x

terms together;

x \times x = x^2

gives the square term.

2Bx^2+(8+AB)x+4A

Now each power of

x

appears once.

3Match the coefficients

#13 Convert to Algebra 6.EE.A.4

Equal for every

x

means equal coefficients:

x^2

with

x^2

,

x

with

x

, constant with constant.

2B=10,\ 8+AB=C,\ 4A=-24

Three equations come out of one.

4Solve the small equations

#13 Convert to Algebra 6.EE.A.4

The first and last give

B

and

A

at once; then the middle one gives

C

.

B=5,\ A=-6,\ C=8+(-6)\times 5=-22

Solve the easy ones first, then use them.

5Add them

#7 Identify Subproblems A-APR.A.1

Add the three constants.

A+B+C=-6+5-22=-23

A

and

B

together give

-1

.

Answer:

-23

4 · Reviewdoes it hold up?

Put the values back: $2x(5x+4)-6(5x+4)=10x^2+8x-30x-24=10x^2-22x-24$ , which matches the right side with $C=-22$ .

Another way: Plug in convenient values of $x$ . With $x=0$ : $4A=-24$ , so $A=-6$ . The left side is $(2x+A)(Bx+4)$ , so its $x^2$ coefficient $2B$ must be $10$ , giving $B=5$ ; then $x=1$ gives $(2-6)(5+4)=-36=10+C-24$ , so $C=-22$ .

Standardsmin grade 9

A-APR.A.1 Polynomials are closed under addition, subtraction, multiplication — Multiplying out both products and collecting the polynomial by powers of $x$.
6.EE.A.4 Identify equivalent expressions — Equal expressions have equal coefficients on like terms.

💡Takeaway. If two polynomials are equal for every

x

, line up their terms and the unknowns fall out one by one.

Match coefficients of like terms · 12 practice problems

Generated variants — 12

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Where next?