Question

second solution
similarly, setting the second factor equal to
$x + 2 = 0 \quad \boldsymbol{\
ightarrow} \quad x = -2$
so, the equation $(x - 1)(x + 2) = 0$ has exactly two solutions: $x = 1$ and $x = -2$.
check these solutions by substituting them into the original equation.
solutions in english & spanish and tutorial videos at collegeprepalgebra.com

exercises within reach®

understanding the zero-factor property in exercises 1–6, determine whether the equation
is written in the correct form to apply the zero-factor property.

1. $x(x-1)=2$
2. $(x-1)(x+1)=1$
3. $(x + 2)+(x - 1)=0$
4. $x(x-3)=0$
5. $(x-1)(x+2)=0$
6. $3(x^{2}+x)=0$

using the zero-factor property in exercises 7–16, use the zero-factor property to solve
the equation.

1. $x(x - 4)=0$
2. $6d(d - 2)=0$
3. $(y - 3)(y + 10)=0$
4. $(s - 7)(s + 4)=0$
5. $25(a + 4)(a - 2)=0$
6. $-8(2 - x)(x + 3)=0$
7. $(3t + 4)(5 - 2t)=0$
8. $(5x - 3)(2x - 8)=0$
9. $-3z(4z + 1)(z - 5)=0$
10. $(y - 39)(2y + 7)(y + 12)=0$

Explanation:

Exercises 1-6: Verify Zero-Factor Property

The Zero-Factor Property states that if $ab=0$, then $a=0$ or $b=0$. We check if each equation is set to 0 with factored terms, or rearrange to match this form.

Step1: Analyze each equation

Check if equation matches $ab=0$ form.
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1. $x(x-1)=2$: Not set to 0 → does not apply
2. $(x-1)(x+1)=1$: Not set to 0 → does not apply
3. $(x+2)+(x-1)=0$: Sum, not product → does not apply
4. $x(x-3)=0$: Product = 0 → applies
5. $(x-1)(x+2)=0$: Product = 0 → applies
6. $3(x^2+x)=0$: Product = 0 → applies

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Exercises 7-16: Solve via Zero-Factor Property

For each equation $ab=0$, set each factor equal to 0 and solve for the variable.

Step1: Set each factor to 0

Solve for the unknown variable.
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1. $x(x-4)=0$

$x=0$ or $x-4=0$ → $x=0$ or $x=4$

1. $6d(d-2)=0$

$6d=0$ or $d-2=0$ → $d=0$ or $d=2$

1. $(y-3)(y+10)=0$

$y-3=0$ or $y+10=0$ → $y=3$ or $y=-10$

1. $(s-7)(s+4)=0$

$s-7=0$ or $s+4=0$ → $s=7$ or $s=-4$

1. $25(a+4)(a-2)=0$

$a+4=0$ or $a-2=0$ → $a=-4$ or $a=2$

1. $-8(2-x)(x+3)=0$

$2-x=0$ or $x+3=0$ → $x=2$ or $x=-3$

1. $(3t+4)(5-2t)=0$

$3t+4=0$ or $5-2t=0$ → $t=-\frac{4}{3}$ or $t=\frac{5}{2}$

1. $(5x-3)(2x-8)=0$

$5x-3=0$ or $2x-8=0$ → $x=\frac{3}{5}$ or $x=4$

1. $-3z(4z+1)(z-5)=0$

$-3z=0$ or $4z+1=0$ or $z-5=0$ → $z=0$, $z=-\frac{1}{4}$, or $z=5$

1. $(y-39)(2y+7)(y+12)=0$

$y-39=0$ or $2y+7=0$ or $y+12=0$ → $y=39$, $y=-\frac{7}{2}$, or $y=-12$

Answer:

Exercises 1-6

1. Does not apply
2. Does not apply
3. Does not apply
4. Applies
5. Applies
6. Applies

Exercises 7-16

1. $x=0$ or $x=4$
2. $d=0$ or $d=2$
3. $y=3$ or $y=-10$
4. $s=7$ or $s=-4$
5. $a=-4$ or $a=2$
6. $x=2$ or $x=-3$
7. $t=-\frac{4}{3}$ or $t=\frac{5}{2}$
8. $x=\frac{3}{5}$ or $x=4$
9. $z=0$, $z=-\frac{1}{4}$, or $z=5$
10. $y=39$, $y=-\frac{7}{2}$, or $y=-12$