ACT Quadratic Factoring: Find the Factors, Check the Roots

Solve quadratic equations by factoring with original worked examples, a leading coefficient greater than one, common mistakes, and four practice questions.

KiboOctober 24, 2026

To solve a quadratic equation by factoring, move all terms to one side so the other side is zero, write the quadratic as a product, and set each variable-containing factor equal to zero. Solve those smaller equations, then substitute each answer into the original equation.

ACT’s Mathematics Standards include factoring simple quadratics and solving quadratic equations in factored form. The examples below are original Kibo practice, not official ACT questions or a prediction of how often a topic will appear.

Why a product equal to zero helps

The zero-product property says that if $AB=0$ , then $A=0$ or $B=0$ , or both. Two nonzero real numbers cannot multiply to zero. This is what turns a factored quadratic into simpler equations. See OpenStax’s explanation of quadratic equations and the zero-product property.

For a quadratic written as $ax^2+bx+c=0$ , the coefficient $a$ must be nonzero. Factoring rewrites the expression; solving finds the values of $x$ that make the equation true. Those values are its solutions, also called roots.

Start with a leading coefficient of one

Solve $x^2-13x+40=0$ . To factor it as $(x+m)(x+n)$ , choose numbers with product $mn=40$ and sum $m+n=-13$ . The numbers $-5$ and $-8$ meet both conditions.

$x^2-13x+40=(x-5)(x-8)$

Multiplying back checks the choice: $x^2-8x-5x+40=x^2-13x+40$ . Because the product equals zero, solve $x-5=0$ or $x-8=0$ . The roots are $x=5$ and $x=8$ .

Check $x=5$ : $25-65+40=0$ .
Check $x=8$ : $64-104+40=0$ .

The factor constants are $-5$ and $-8$ ; the roots are $5$ and $8$ . Set each factor equal to zero instead of copying the signs inside the parentheses. If that last step is difficult, review the linear-equation balance method.

When the leading coefficient is larger than one

Solve $6x^2+x-12=0$ . A candidate of the form $(3x+p)(2x+q)$ has constant term $pq$ and middle coefficient $3q+2p$ . Both must match the original expression.

Choose $p=-4$ and $q=3$ : the constant product is $(-4)(3)=-12$ , and the middle coefficient is $3(3)+2(-4)=9-8=1$ . Therefore:

$6x^2+x-12=(3x-4)(2x+3)$

The expansion is $6x^2+9x-8x-12$ , so the middle terms combine correctly. Set $3x-4=0$ or $2x+3=0$ . This gives $x=\frac{4}{3}$ or $x=-\frac{3}{2}$ .

For $x=\frac{4}{3}$ : $6\left(\frac{4}{3}\right)^2+\frac{4}{3}-12=\frac{32}{3}+\frac{4}{3}-\frac{36}{3}=0$ .
For $x=-\frac{3}{2}$ : $6\left(-\frac{3}{2}\right)^2-\frac{3}{2}-12=\frac{27}{2}-\frac{3}{2}-\frac{24}{2}=0$ .

Matching only the first and last terms is insufficient. For example, $(3x+4)(2x-3)$ expands to $6x^2-x-12$ , which has the wrong middle sign for this problem. Always check the cross terms.

Put zero on one side before using the property

Solve $x^2+3x=54$ . Subtract $54$ from both sides, giving $x^2+3x-54=0$ . Since $9(-6)=-54$ and $9-6=3$ , the equation becomes $(x+9)(x-6)=0$ .

The roots are $x=-9$ and $x=6$ . Check the original equation: $(-9)^2+3(-9)=81-27=54$ , and $6^2+3(6)=36+18=54$ .

A product with a nonzero result needs the same care. From $(x-5)(x+2)=18$ , you cannot conclude that either factor is zero. Expand and rearrange to obtain $x^2-3x-28=0$ , then factor as $(x-7)(x+4)=0$ .

The answers are $x=7$ and $x=-4$ . Substituting into the original product gives $(7-5)(7+2)=2(9)=18$ and $(-4-5)(-4+2)=(-9)(-2)=18$ .

Keep a possible zero solution

Solve $5x^2-35x=0$ by taking out the common factor: $5x(x-7)=0$ . The constant factor $5$ is never zero, so $x=0$ or $x-7=0$ . Both $0$ and $7$ are solutions.

Dividing the original equation by $x$ would assume $x\neq 0$ and lose a valid answer. Factoring preserves it. The checks are $5(0)^2-35(0)=0$ and $5(7)^2-35(7)=245-245=0$ .

Try four original practice questions

Find all real solutions. Show the factorization and check the original equation before opening the answers.

$x^2+5x-66=0$
$8x^2-2x-3=0$
$3x^2=27x$
$9x^2-30x+25=0$

Show answers and checks

$(x-6)(x+11)=0$ , so $x=6$ or $x=-11$ . The original expression gives $36+30-66=0$ and $121-55-66=0$ .
$(4x-3)(2x+1)=0$ , so $x=\frac{3}{4}$ or $x=-\frac{1}{2}$ . Check: $\frac{9}{2}-\frac{3}{2}-3=0$ and $2+1-3=0$ .
Move all terms left: $3x^2-27x=3x(x-9)=0$ . The roots are $0$ and $9$ . In the original equation, both sides equal $0$ at $x=0$ , and both equal $243$ at $x=9$ .
$(3x-5)^2=0$ , so the only distinct root is $x=\frac{5}{3}$ . The repeated factor gives the same root twice. Check: $25-50+25=0$ .

Know when to change methods

Some quadratics do not factor using integer constants. That does not mean they have no real solutions. For example, $x^2-11=0$ has real roots $x=\sqrt{11}$ and $x=-\sqrt{11}$ , since each squares to $11$ . OpenStax’s overview of quadratic methods also covers square roots, completing the square, and the quadratic formula. Use another appropriate method when a simple factorization is unavailable.

For your next attempt, check four things: zero on one side, correct expansion, every factor considered, and every proposed root substituted into the original equation. If a mistake repeats, record its first incorrect step in your practice error log. For sign handling, revisit the negative-sign and exponent examples.

Try a free ACT practice question with Kibo, then save your reasoning and choose one skill to review.

Prepared with AI assistance. The worked problems and practice questions are original; their expansions and solutions were checked separately with exact arithmetic. No human review or score improvement is claimed. Sources checked October 24, 2026. This guide is not affiliated with or endorsed by ACT.