Question

second solution
similarly, setting the second factor equal to
$x + 2 = 0 \implies x = -2$
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 Form

Step1: Recall Zero-Factor Property

A equation is in correct Zero-Factor Property form if it is written as $\text{Factor}_1 \times \text{Factor}_2 \times ... = 0$.

Step2: Analyze each equation

1. $x(x-1)=2$: Right-hand side is 2, not 0 → Not correct form.
2. $(x-1)(x+1)=1$: Right-hand side is 1, not 0 → Not correct form.
3. $(x+2)+(x-1)=0$: Uses addition, not multiplication of factors → Not correct form.
4. $x(x-3)=0$: Product of factors equals 0 → Correct form.
5. $(x-1)(x+2)=0$: Product of factors equals 0 → Correct form.
6. $3(x^2+x)=0$: Product of a constant and a factor equals 0 → Correct form.

Rule: Zero-Factor Property

If $ab=0$, then $a=0$ or $b=0$. Set each factor equal to 0 and solve for the variable.

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Exercise7: $x(x-4)=0$

Step1: Set each factor to 0

$x=0$ or $x-4=0$

Step2: Solve for $x$

$x=0$ or $x=4$

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Exercise8: $6d(d-2)=0$

Step1: Set each factor to 0

$6d=0$ or $d-2=0$

Step2: Solve for $d$

$d=0$ or $d=2$

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Exercise9: $(y-3)(y+10)=0$

Step1: Set each factor to 0

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

Step2: Solve for $y$

$y=3$ or $y=-10$

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Exercise10: $(s-7)(s+4)=0$

Step1: Set each factor to 0

$s-7=0$ or $s+4=0$

Step2: Solve for $s$

$s=7$ or $s=-4$

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Exercise11: $25(a+4)(a-2)=0$

Step1: Set each factor to 0

$a+4=0$ or $a-2=0$ (25≠0, so it does not yield a solution)

Step2: Solve for $a$

$a=-4$ or $a=2$

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Exercise12: $-8(2-x)(x+3)=0$

Step1: Set each factor to 0

$2-x=0$ or $x+3=0$ (-8≠0, so it does not yield a solution)

Step2: Solve for $x$

$x=2$ or $x=-3$

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Exercise13: $(3t+4)(5-2t)=0$

Step1: Set each factor to 0

$3t+4=0$ or $5-2t=0$

Step2: Solve for $t$

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

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Exercise14: $(5x-3)(2x-8)=0$

Step1: Set each factor to 0

$5x-3=0$ or $2x-8=0$

Step2: Solve for $x$

$x=\frac{3}{5}$ or $x=4$

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Exercise15: $-3z(4z+1)(z-5)=0$

Step1: Set each factor to 0

$-3z=0$ or $4z+1=0$ or $z-5=0$

Step2: Solve for $z$

$z=0$ or $z=-\frac{1}{4}$ or $z=5$

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Exercise16: $(y-39)(2y+7)(y+12)=0$

Step1: Set each factor to 0

$y-39=0$ or $2y+7=0$ or $y+12=0$

Step2: Solve for $y$

$y=39$ or $y=-\frac{7}{2}$ or $y=-12$

Answer:

1. Not in correct form
2. Not in correct form
3. Not in correct form
4. In correct form
5. In correct form
6. In correct form

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Exercises 7-16: Solve via Zero-Factor Property