Question

%7D%7B4%20%2B%203x%20-%20%7Bx%7D%5E%7B2%7D%20%7D%20" id="TexFormula1" title=" \frac{ {9x}^{2} - {(x}^{2} - 4) {}^{2} }{4 + 3x - {x}^{2} } " alt=" \frac{ {9x}^{2} - {(x}^{2} - 4) {}^{2} }{4 + 3x - {x}^{2} } " align="absmiddle" class="latex-formula">
pls help me need help asap

Answer 1

Answer:

The answer is

x² + 3x - 4

Step-by-step explanation:

$\frac{9 {x}^{2} - ( { {x}^{2} - 4})^{2} }{4 + 3x - {x}^{2} }$

To solve the expression first factorize both the numerator and the denominator

For the numerator

9x² - ( x² - 4)²

Expand the terms in the bracket using the formula

( a - b)² = a² - 2ab + b²

(x² - 4) = x⁴ - 8x² + 16

So we have

9x² - (x⁴ - 8x² + 16)

9x² - x⁴ + 8x² - 16

- x⁴ + 17x² - 16

Factorize

that's

(x² - 16)(-x² + 1)

Using the formula

a² - b² = ( a + b)(a - b)

We have

(x² - 16)(-x² + 1) = (x + 4)(x - 4)( 1 - x)(1 + x)

For the denominator

- x² + 3x + 4

Write 3x as a difference

- x² + 4x - x + 4

Factorize

That's

- ( x - 4)(x + 1)

So we now have

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

Simplify

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

Reduce the expression by x + 1

That's

-( x + 4)( 1 - x)

Multiply the terms

We have the final answer as

x² + 3x - 4

Hope this helps you

Answer 2

Answer:

${ x^2+3x-4}$

Step-by-step explanation:

Factor top and bottom.

The numerator is a difference of two squares, and the denominator is a quadratic.

$\frac{ {9x}^{2} - {(x}^{2} - 4)^{2} }{4 + 3x - {x}^{2} }$

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

= $\frac{ (x-1)(x+4) (1+x)(4-x) }{(1+x)(4-x)}$

If x does not equal -1 and does not equal 4, we can cancel the common factors in italics to give

= ${ (x-1)(x+4)}$

= ${ x^2+3x-4}$