Answer:
x³/2
Step-by-step explanation:
Factor the top and bottom completely
x³(x + 2) · 2(x+1)(x-1)
4(x+2)(x-1) (x+1)
Cancel out any matches on the top and bottom
x³/2
Answer:
[tex]\textsf{1)} \quad \dfrac{x}{2}[/tex]
[tex]\textsf{2)}\quad \dfrac{2}{3}[/tex]
Step-by-step explanation:
Given expression:
[tex]\dfrac{x^4+2x^3}{4x^4+4x^3-8x^2}\cdot \dfrac{2x^2-2}{x+1}[/tex]
To simplify the given expression, begin by factoring the numerators and denominators, where applicable.
To factor the numerator of the first fraction, factor out the common term x³:
[tex]x^4+2x^3=x^3(x+2)[/tex]
To factor the denominator of the first fraction, begin by factoring out the common term 4x², then factor the quadratic inside the parentheses:
[tex]\begin{aligned}4x^4+4x^3-8x^2&=4x^2(x^2+x-2)\\&=4x^2(x^2+2x-x-2)\\&=4x^2(x(x+2)-1(x+2))\\&=4x^2(x-1)(x+2)\end{aligned}[/tex]
To factor the numerator of the second fraction, begin by factoring out the common term 2, then factor the quadratic inside the parentheses:
[tex]\begin{aligned}2x^2-2&=2(x^2-1)\\&=2(x^2+x-x-1)\\&=2(x(x+1)-1(x+1))\\&=2(x+1)(x-1)\end{aligned}[/tex]
Therefore, the expression with factored numerators and denominators is:
[tex]\dfrac{x^3(x+2)}{4x^2(x-1)(x+2)}\cdot \dfrac{2(x+1)(x-1)}{x+1}[/tex]
Cancel the common factors of each fraction:
[tex]\dfrac{x}{4(x-1)}\cdot 2(x-1)[/tex]
Cross-cancel the common factor (x - 2):
[tex]\dfrac{x}{4} \cdot 2[/tex]
[tex]\dfrac{2x}{4}[/tex]
Cancel the common factor 2:
[tex]\dfrac{x}{2}[/tex]
Therefore:
[tex]\dfrac{x^4+2x^3}{4x^4+4x^3-8x^2}\cdot \dfrac{2x^2-2}{x+1}=\boxed{\dfrac{x}{2}}[/tex]
[tex]\hrulefill[/tex]
Given expression:
[tex]\dfrac{x+10}{x^2+20x+100}\cdot \dfrac{6x}{3x}\cdot \dfrac{x^2+3x-70}{3x-21}[/tex]
To simplify the given expression, begin by factoring the numerators and denominators, where applicable.
Factor the denominator of the first fraction:
[tex]\begin{aligned}x^2+20x+100&=x^2+10x+10x+100\\&=x(x+10)+10(x+10)\\&=(x+10)(x+10)\end{aligned}[/tex]
Factor the numerator of the third fraction:
[tex]\begin{aligned}x^2+3x-70&=x^2+10x-7x-70\\&=x(x+10)-7(x+10)\\&=(x+10)(x-7)\end{aligned}[/tex]
Factor the denominator of the third fraction:
[tex]3x-21=3(x-7)[/tex]
Therefore, the expression with factored numerators and denominators is:
[tex]\dfrac{x+10}{(x+10)(x+10)}\cdot \dfrac{6x}{3x}\cdot \dfrac{(x+10)(x-7)}{3(x-7)}[/tex]
Cancel the common factors of each fraction:
[tex]\dfrac{1}{x+10}\cdot 2\cdot \dfrac{x+10}{3}[/tex]
Cross-cancel the common factor (x + 10):
[tex]2 \cdot \dfrac{1}{3}[/tex]
[tex]\dfrac{2}{3}[/tex]
Therefore:
[tex]\dfrac{x+10}{x^2+20x+100}\cdot \dfrac{6x}{3x}\cdot \dfrac{x^2+3x-70}{3x-21}=\boxed{\dfrac{2}{3}}[/tex]
