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3 years ago

2x^2+5x+3/x^2-3x-4 divided by 4x^2+2x-6/x^2-8x+16

Mathematics

1 answer:

Masja [62]3 years ago

8 0

Answer: $\frac{x-4}{2x-2}$

Step-by-step explanation:

$\frac{2x^2+5x+3}{x^2-3x-4}$ divided by $\frac{4x^2+2x-6}{x^2-8x+16}$ is the same thing as multiplying $\frac{2x^2+5x+3}{x^2-3x-4}$ by $\frac{x^2-8x+16}{4x^2+2x-6}$ .

The equation we get is:

$\frac{2x^2+5x+3}{x^2-3x-4}*\frac{x^2-8x+16}{4x^2+2x-6}$

With this equation, we can factor each equation:

$\frac{(x+1)(2x+3)}{(x+1)(x-4)} *\frac{(x-4)(x-4)}{2(x-1)(2x+3)}$

We can cancel out like terms since they would be dividing each other:

$\frac{x-4}{2x-2}$

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Step-by-step explanation:

The picture of the question in the attached figure

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Darina [25.2K]

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