Factoring quadratics: Algebra 1 Regents Practice

Factor completely: $x^2 + 7x + 12$

• (A) $(x+3)(x+4)$ ✓
• (B) $(x+2)(x+6)$
• (C) $(x-3)(x-4)$
• (D) $(x+1)(x+12)$

For a trinomial $x^2+bx+c$ with leading coefficient 1, find two numbers that multiply to $c$ and add to $b$.

Factor completely: $6x^2 + 9x$

• (A) $x(6x+9)$
• (B) $3(2x^2+3x)$
• (C) $3x(2x+9)$
• (D) $3x(2x+3)$ ✓

Factor out the greatest common factor of every term, dividing each term by the full GCF (numbers and variables).

Factor the expression completely: $2x^3 - 8x$

Factoring completely means removing the GCF first, then continuing to factor what remains. Here the remaining binomial $x^2 - 4$ is a difference of perfect squares, which factors into conjugate binomials.

Solve the equation $x^2 + 3x - 28 = 0$ algebraically by factoring. Show your work.

The quadratic factors into two binomials because $7 \cdot (-4) = -28$ and $7 + (-4) = 3$. By the zero product property, a product equals zero only when at least one factor equals zero, giving the two solutions $x = -7$ and $x = 4$.

What are the zeros of the function $f(x) = x(x - 5)(3x + 6)$ ?

• (A) $0,\ 5,\ \text{and } 2$
• (B) $5 \text{ and } -2 \text{, only}$
• (C) $0,\ -5,\ \text{and } 2$
• (D) $0,\ 5,\ \text{and } -2$ ✓

A function in fully-factored form has a zero wherever any single factor equals zero. Set each factor to 0 and solve, don't forget the bare x.

Factored completely, $x^2 - 36$ is equivalent to

• (A) $(x-9)(x+4)$
• (B) $(x-36)(x+1)$
• (C) $(x-6)(x+6)$ ✓
• (D) $(x-6)^2$

a² − b² always factors to (a − b)(a + b), with no middle term.

Factor completely: $4x^2 - 25$

• (A) $(2x-5)^2$
• (B) $(4x-5)(4x+5)$
• (C) $(2x-25)(2x+25)$
• (D) $(2x-5)(2x+5)$ ✓

A difference of squares $a^2-b^2$ factors as $(a-b)(a+b)$; take the square root of each term to find $a$ and $b$.

Factor completely: $x^2 - 3x - 10$

• (A) $(x-5)(x+2)$ ✓
• (B) $(x+5)(x-2)$
• (C) $(x+5)(x+2)$
• (D) $(x-5)(x-2)$

When the constant is negative, the two numbers have opposite signs, and the number with the larger absolute value carries the sign of the middle term.

Factored completely, the expression $x^2 - 5x - 14$ is equivalent to

• (A) $(x - 7)(x + 2)$ ✓
• (B) $(x + 7)(x + 2)$
• (C) $(x + 7)(x - 2)$
• (D) $(x - 7)(x - 2)$

For a trinomial with leading coefficient $1$, find two numbers that multiply to the constant term $-14$ and add to the middle coefficient $-5$. Those are $-7$ and $2$, giving $(x - 7)(x + 2)$. Swapping the signs changes the middle term to $+5x$ or the product to $+14$, which the distractors reflect.

Solve the equation $x^2 = 6x + 27$ algebraically. Show all work.

Setting the quadratic equal to zero lets you use factoring and the zero product property, which says that if a product equals zero at least one factor must equal zero. Solving each factor gives the two roots, and both check when substituted back into the original equation.

The length of a rectangle is 5 feet more than its width. The area of the rectangle is 84 square feet. Determine the width and the length of the rectangle, in feet, algebraically.

Letting the width be $w$, the length is $w + 5$, so the area gives the quadratic $w^2 + 5w - 84 = 0$. Factoring produces $w = 7$ or $w = -12$; since a width cannot be negative, the width is 7 feet and the length is 12 feet.

Which binomial is a factor of $x^2 - 3x - 40$?

• (A) $x - 5$
• (B) $x + 8$
• (C) $x + 5$ ✓
• (D) $x - 40$

The factor pair of $-40$ that adds to $-3$ is $-8$ and $5$, so the factors are $(x - 8)$ and $(x + 5)$. Of the listed choices, only $x + 5$ appears, so it is the correct factor.

Which expression is equivalent to $x^2 - 5x - 24$?

• (A) $(x - 6)(x + 4)$
• (B) $(x + 8)(x - 3)$
• (C) $(x - 8)(x + 3)$ ✓
• (D) $(x - 12)(x + 2)$

To factor a trinomial with leading coefficient 1, find two numbers whose product is the constant term, $-24$, and whose sum is the coefficient of $x$, $-5$. The pair $-8$ and $3$ works, giving $(x-8)(x+3)$. The distractors use sign errors or factor pairs of $-24$ that do not sum to $-5$.

Solve the equation $x^2-5x-24=0$ algebraically for all values of $x$.

The quadratic factors because $-8$ and $3$ multiply to $-24$ and add to $-5$. By the zero product property, a product equals zero only when at least one factor is zero, so the solutions are $x=8$ and $x=-3$.

Factor completely: $3x^3-12x$

Factoring completely means first removing the greatest common factor, then factoring any remaining factorable expression. After factoring out $3x$, the binomial $x^2-4$ is a difference of two perfect squares, so the complete factorization is $3x(x+2)(x-2)$.

Which expression is a factor of $x^2 - 5x - 24$?

• (A) $x - 3$
• (B) $x - 4$
• (C) $x - 8$ ✓
• (D) $x + 8$

To factor $x^2 - 5x - 24$, find two numbers whose product is $-24$ and whose sum is $-5$. Those numbers are $-8$ and $3$, so the factorization is $(x - 8)(x + 3)$, making $x - 8$ a factor. Choosing $x + 8$ or $x - 3$ comes from mixing up the signs.

What is the solution set of the equation $x^2 - 5x - 24 = 0$?

• (A) $\{-3, 8\}$ ✓
• (B) $\{-3, -8\}$
• (C) $\{3, -8\}$
• (D) $\{3, 8\}$

Factoring requires two numbers with product $-24$ and sum $-5$, which are $-8$ and $3$, giving factors $(x-8)(x+3)$. Setting each factor equal to zero gives $x = 8$ and $x = -3$. The distractors come from sign errors when converting factors to roots.

Solve the equation $x^2 - 5x - 24 = 0$ algebraically by factoring.

The quadratic factors using two numbers whose product is $-24$ and whose sum is $-5$, namely $-8$ and $3$. By the zero product property, each factor set equal to zero gives a solution, so $x = 8$ or $x = -3$.

Factor completely: $3x^2 + 10x + 8$

• (A) $(3x-4)(x-2)$
• (B) $(3x+2)(x+4)$
• (C) $(3x+8)(x+1)$
• (D) $(3x+4)(x+2)$ ✓

When the leading coefficient is not 1, split the middle term using two numbers that multiply to $a\cdot c$ and add to $b$, then factor by grouping.

What are the solutions to $x^2 + 2x - 15 = 0$?

• (A) $x=5$ and $x=3$
• (B) $x=5$ and $x=-3$
• (C) $x=-5$ and $x=3$ ✓
• (D) $x=-5$ and $x=-3$

Factor, then set each factor equal to zero (zero product property); each root is the value that makes its factor zero, which is the opposite of the number inside.

Which expression is equivalent to $3x^2 - 10x - 8$?

• (A) $(3x - 2)(x + 4)$
• (B) $(3x + 2)(x - 4)$ ✓
• (C) $(3x + 4)(x - 2)$
• (D) $(3x - 4)(x + 2)$

Factoring a trinomial with a leading coefficient of 3 requires splitting the middle term using two numbers that multiply to $-24$ and add to $-10$, namely $-12$ and $2$. Checking by multiplication, $(3x+2)(x-4) = 3x^2 - 12x + 2x - 8 = 3x^2 - 10x - 8$. The distractors give the wrong middle term or the wrong signs when expanded.

Solve the equation $2x^2 - 5x - 3 = 0$ algebraically for all values of $x$. Justify your solution by showing your work.

Because the leading coefficient is not 1, the quadratic is factored by splitting the middle term using two numbers whose product is $ac = -6$ and whose sum is $b = -5$. The zero product property then says each factor can equal zero, giving the two solutions $x = 3$ and $x = -\frac{1}{2}$, both of which check in the original equation.