Question

algebra 1.5
chp 11 test {part 2 } modified
name dylan h.
return for grading
must show your work !!!
factor if possible, state the restrictions, and then simplify the expression.

1. $\frac{10x^{2}}{15x^{3}}$

restrictions: $x \
eq$

1. $\frac{x^{2}+4x-5}{2-2x}$

check for standard form and gcf.
restrictions: $x \
eq$

1. $\frac{x^{2}+4x-21}{2x^{2}+11x-21}$

restrictions: $x \
eq$
cornillew/needs more time

• need to

Explanation:

Problem 1:

Step1: Identify GCF of terms

GCF of $10x^2$ and $15x^3$ is $5x^2$.

Step2: Cancel common factors

$\frac{10x^2}{15x^3} = \frac{5x^2 \cdot 2}{5x^2 \cdot 3x} = \frac{2}{3x}$

Step3: Find restrictions

Denominator $15x^3
eq 0 \implies x
eq 0$

Problem 2:

Step1: Factor numerator and denominator

Numerator: $x^2+4x-5=(x+5)(x-1)$
Denominator: $2-2x=-2(x-1)$

Step2: Cancel common factors

$\frac{(x+5)(x-1)}{-2(x-1)} = -\frac{x+5}{2}$

Step3: Find restrictions

Denominator $2-2x
eq 0 \implies x
eq 1$

Problem 3:

Step1: Factor numerator and denominator

Numerator: $x^2+4x-21=(x+7)(x-3)$
Denominator: $2x^2+11x-21=(2x-3)(x+7)$

Step2: Cancel common factors

$\frac{(x+7)(x-3)}{(2x-3)(x+7)} = \frac{x-3}{2x-3}$

Step3: Find restrictions

Denominator $2x^2+11x-21
eq 0 \implies 2x-3
eq 0$ and $x+7
eq 0 \implies x
eq \frac{3}{2}, x
eq -7$

Answer:

1. Simplified: $\frac{2}{3x}$, Restrictions: $x

eq 0$

1. Simplified: $-\frac{x+5}{2}$, Restrictions: $x

eq 1$

1. Simplified: $\frac{x-3}{2x-3}$, Restrictions: $x

eq \frac{3}{2}, x
eq -7$