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Factoring by Grouping: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 A cubic in four terms

Factor $3x^3-21x^2-5x+35$ by grouping, and check your factorization by expanding it.
1. Hint 1
  Grouping only finishes when both pairs leave the same binomial behind, so watch what each pair leaves.
2. Hint 2
  The second pair leads with a negative term; choose its common factor so that its binomial matches the first pair's.
Answer

$(x-7)(3x^2-5)$ .

Full solution

Pair the first two terms and the last two terms.
The first pair gives $3x^2(x-7)$ .
In the second pair, taking out $+5$ would leave $5(7-x)$ , the opposite of $x-7$ , so take out $-5$ instead and get $-5(x-7)$ .
Both pairs now carry $x-7$ , so that binomial factors out:
$(x-7)(3x^2-5).$
Expanding gives $3x^3-5x-21x^2+35$ , which reorders to the original polynomial.
The factor $3x^2-5$ has no factorization with integer coefficients, since neither $3$ nor $5$ is a perfect square, so this is complete.

Answer

$(x-7)(3x^2-5)$ .

Key idea

A pair that leads with a negative term needs a negative common factor so its binomial matches the other pair's.
Problem 1: Mark complete
Problem 2 A shared binomial in letters

Write $A(2x-5)+B(5-2x)$ as a product of two binomials for all real $A,B,x$ .
1. Hint 1
  Nothing can be pulled out until both groups carry the same binomial, so compare the two parenthesized expressions first.
2. Hint 2
  Factor a negative from the reversed expression before extracting the common binomial.
Answer

$(A-B)(2x-5)$ .

Full solution

The reversal gives
$5-2x=-(2x-5).$
Thus the expression becomes $A(2x-5)-B(2x-5)$ .
The common binomial factors out to give
$(A-B)(2x-5).$
Distributing it reproduces both original groups with their signs intact.

Answer

$(A-B)(2x-5)$ .

Key idea

Opposite binomials become a shared factor when one group contributes a negative multiplier.
Problem 2: Mark complete
Problem 3 A harder trinomial

Factor $10x^2-9x-7$ by the AC method, and check your factorization by expanding it.
1. Hint 1
  The middle coefficient splits into two pieces whose product is $ac$ , the first coefficient times the constant.
2. Hint 2
  Look for two integers with product $-70$ and sum $-9$ , then split the middle term and group the four terms.
3. Hint 3
  When a pair factors down to a bare binomial, keep the $1$ in front of it; it becomes a coefficient in the answer.
Answer

$(5x-7)(2x+1)$ .

Full solution

Here $a=10$ , $b=-9$ and $c=-7$ , so the product to aim for is $ac=-70$ .
Two integers with product $-70$ and sum $-9$ are $-14$ and $5$ .
Split the middle term into $-14x+5x$ :
$10x^2-14x+5x-7.$
The first pair gives $2x(5x-7)$ , and the second gives $1(5x-7)$ , keeping the $1$ .
Factoring out the shared binomial leaves
$(5x-7)(2x+1).$
Expanding gives $10x^2+5x-14x-7$ , which collects to $10x^2-9x-7$ .

Answer

$(5x-7)(2x+1)$ .

Key idea

Splitting the middle term with the pair whose product is $ac$ turns a trinomial into a grouping problem.
Problem 3: Mark complete
Problem 4 An equality of totals

Find every real solution of $3x^3+15x=4x^2+20$ , showing a grouping of the four terms after moving them to one side.
1. Hint 1
  Place all four terms on the same side before looking for shared factors.
2. Hint 2
  Once the polynomial is written as a product, ask which of its factors can be zero for a real $x$ .
Answer

$x=\frac43$ , the only real solution.

Full solution

Moving every term to the left gives
$3x^3-4x^2+15x-20=0.$
The first pair gives $x^2(3x-4)$ , and the second gives $5(3x-4)$ .
Factoring out the shared binomial produces
$(3x-4)(x^2+5)=0.$
A product is zero only when one of its factors is zero.
The factor $x^2+5$ is never zero for a real $x$ , because $x^2$ is never negative, so $x^2+5$ is at least $5$ .
The only real solution therefore comes from $3x-4=0$ , which gives $x=\frac43$ .
Substituting that value gives $\frac{64}{9}+20$ on each side of the original equation.

Answer

$x=\frac43$ , the only real solution.

Key idea

Grouping turns an equation into a product, and only the factors that can vanish contribute real solutions.
Problem 4: Mark complete
Problem 5 A missing coefficient

The polynomial $6x^3+Cx^2-5x-20$ has the factor $x+4$ . Find $C$ and give its complete factorization with integer coefficients.
1. Hint 1
  The last two terms already have the required binomial factor after removing their common factor.
2. Hint 2
  Determine which coefficient lets the first two terms share the same binomial, then inspect the remaining factor.
Answer

$C=24$ ; $(x+4)(6x^2-5)$ .

Full solution

The last pair is
$-5x-20=-5(x+4).$
For the two pairs to leave the same binomial, the first pair $x^2(6x+C)$ must be a multiple of $x+4$ as well, and a linear expression that is a multiple of $x+4$ is just a constant times $x+4$ .
Matching the leading $6$ makes that constant $6$ , so $6x+C=6(x+4)$ and
$C=24.$
Substituting $x=-4$ into $6x^3+24x^2-5x-20$ gives $0$ , confirming the factor.
The groups combine to
$(x+4)(6x^2-5).$
The factor $6x^2-5$ has no factorization with integer coefficients, since neither $6$ nor $5$ is a perfect square, so this is complete.

Answer

$C=24$ ; $(x+4)(6x^2-5)$ .

Key idea

Matching the binomial one pair leaves can pin down a missing coefficient in the other pair.
Problem 5: Mark complete
Problem 6 Two ways to pair

For $6u^3+10u^2-15u-25$ , grouping $6u^3-25$ together leaves no shared binomial with the remaining pair. Give two different pairings that do work and the resulting product with integer coefficients.
1. Hint 1
  A pairing works only when both pairs leave the same binomial, so look for pairs whose two terms share a factor.
2. Hint 2
  There are three ways to split four terms into two pairs, and the statement has already ruled one of them out.
3. Hint 3
  Factor each pair carefully, taking a negative out when needed so the binomials match.
Answer

Pairings: $(6u^3+10u^2)+(-15u-25)$ and $(6u^3-15u)+(10u^2-25)$ ; product $(3u+5)(2u^2-5)$ .

Full solution

The first working pairing becomes
$2u^2(3u+5)-5(3u+5),$
giving $(3u+5)(2u^2-5)$ .
The other becomes
$3u(2u^2-5)+5(2u^2-5),$
giving the same product.
Expanding either product gives $6u^3+10u^2-15u-25$ .
Thus the initially unsuccessful pairing did not rule out grouping.

Answer

Pairings: $(6u^3+10u^2)+(-15u-25)$ and $(6u^3-15u)+(10u^2-25)$ ; product $(3u+5)(2u^2-5)$ .

Key idea

Different pairings can expose different common binomials while producing the same complete product.
Problem 6: Mark complete
Problem 7 A divided quantity

For $x\neq-\frac52$ , a quantity $4x^2+4x-15$ is divided by $2x+5$ . Simplify the quotient by showing an AC split and grouping, and find its value at $x=3$ .
1. Hint 1
  The split coefficients must have product $-60$ and sum $4$ .
2. Hint 2
  After grouping, cancel the denominator factor, which is nonzero on the given domain.
Answer

Quotient $2x-3$ ; value at $x=3$ : $3$ .

Full solution

The AC pair is $10,-6$ .
Split the polynomial as $4x^2+10x-6x-15$ and group:
$2x(2x+5)-3(2x+5).$
The result is $(2x+5)(2x-3)$ .
The condition makes $2x+5$ nonzero, so the shared factor cancels and the quotient is $2x-3$ .
At $x=3$ , this is $3$ ; the original quotient is $33/11=3$ .

Answer

Quotient $2x-3$ ; value at $x=3$ : $3$ .

Key idea

An AC split can expose a nonzero denominator factor and simplify a quotient.
Problem 7: Mark complete
Problem 8 A constant rule

For positive real numbers $p,q$ , a student claims that choosing $r=p+q$ makes $x^3+px^2+qx+r$ group into $(x+p)(x^2+q)$ . Is the rule correct? State the condition on $r$ that actually makes that product an identity.
1. Hint 1
  Expand the proposed product and compare all four coefficients.
2. Hint 2
  Multiply $x+p$ by each term of $x^2+q$ , then compare the constant term of that expansion with $r$ .
Answer

False in general; the required condition is $r=pq$ .

Full solution

The product expands as $x^3+px^2+qx+pq$ .
Therefore matching the given constant requires
$r=pq.$
The suggested rule fails, for example, when $p=2,q=3$ : it supplies $r=5$ , but the product has constant $6$ .
Some choices may have $pq=p+q$ , but the sum is not the general rule.

Answer

False in general; the required condition is $r=pq$ .

Key idea

The shared-binomial structure of a cubic fixes its constant as a product, not as a sum.
Problem 8: Mark complete
Problem 9 The order of a split

For $12x^2+x-6$ , the AC pair is $9,-8$ . Jules claims that either order of these two middle terms leaves the same shared binomial once the pairs are factored. Is the claim correct? Show both orders.
1. Hint 1
  The shared binomial is whatever both pairs leave behind, so a claim about it is settled by carrying out both groupings.
2. Hint 2
  Write the middle term as $9x-8x$ and then as $-8x+9x$ .
3. Hint 3
  Factor the adjacent pairs in each version, keeping any negative common factor.
Answer

No; the first order leaves $4x+3$ and the second leaves $3x-2$ , though both orders give the factorization $(4x+3)(3x-2)$ .

Full solution

With $9x$ first, the four terms are $12x^2+9x-8x-6$ .
The first pair gives $3x(4x+3)$ and the second gives $-2(4x+3)$ , so the shared binomial is $4x+3$ and the product is
$(4x+3)(3x-2).$

With $-8x$ first, the four terms are $12x^2-8x+9x-6$ .
Now the first pair gives $4x(3x-2)$ and the second gives $3(3x-2)$ , so the shared binomial is $3x-2$ instead.
Its product is $(3x-2)(4x+3)$ , the same two factors in the other order.
Both orders expand to $12x^2+x-6$ , so Jules is wrong about the shared binomial even though the final factorization does not depend on the order.

Answer

No; the first order leaves $4x+3$ and the second leaves $3x-2$ , though both orders give the factorization $(4x+3)(3x-2)$ .

Key idea

Reversing a valid AC split can change the visible shared binomial while preserving the final factors.
Problem 9: Mark complete
Problem 10 Two cubic designs

Create two four-term cubic polynomials with integer coefficients whose $x^3$ coefficient is $1$ . One must group into three linear factors with a repeated factor, and the other must group into three distinct linear factors. Give each expanded polynomial and its complete product, and show the grouping that produces each product.
1. Hint 1
  Begin with a product of a linear binomial and a difference of squares.
2. Hint 2
  For the repeated case, make the linear binomial match one of the factors of the difference; choose a different one for the distinct case.
Answer

Examples: $x^3+3x^2-9x-27=(x+3)^2(x-3)$ ; $x^3+x^2-4x-4=(x+1)(x-2)(x+2)$ .

Full solution

For the repeated case, choose $(x+3)(x^2-9)$ , which expands to $x^3+3x^2-9x-27$ .
Grouping its four terms gives
$x^2(x+3)-9(x+3).$
Pulling out the shared $x+3$ and then factoring the difference of squares gives
$(x+3)^2(x-3).$

For the distinct case, choose $(x+1)(x^2-4)$ , which expands to $x^3+x^2-4x-4$ .
Grouping gives
$x^2(x+1)-4(x+1),$
and factoring the difference of squares gives
$(x+1)(x-2)(x+2).$
Its three binomials are all different.
Any pair of examples meeting the stated conditions is valid.

Answer

Examples: $x^3+3x^2-9x-27=(x+3)^2(x-3)$ ; $x^3+x^2-4x-4=(x+1)(x-2)(x+2)$ .

Key idea

Designing a grouping product allows control over whether its final linear factors repeat.
Problem 10: Mark complete

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