How do you factor $- 7 x^{2} + 175$ $-7x^2+175$ ?

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How do you factor $- 7 x^{2} + 175$ $-7x^2+175$ ?

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Answer 1 · 2018-05-12T05:32:32

Prime factorize them
$- 7 x^{2} = - 1 \times 7 \times x \times x$ $-7x^2=-1xx7xx x xx x$
$175 = 5 \times 5 \times 7$ $175=5xx5xx7$
As you can see... both have 7 as common
$7 (- x^{2} + 25)$ $7(-x^2+25)$
Leaving it like this will not complete the answer
Notice
$7 (25 - x^{2}) = 7 (5^{2} - x^{2})$ $7(25-x^2)=7(5^2-x^2)$

This is a law of exponents
$a^{2} - b^{2} = (a + b) (a - b)$ $a^2-b^2=(a+b)(a-b)$

Here... $5 = a, x = b$ $5=a, x=b$
$7 (5^{2} - x^{2}) = 7 (5 + x) (5 - x)$ $7(5^2-x^2)=7(5+x)(5-x)$

Answer 2 · 2018-05-12T05:33:17

$- 7 (x - 5) (x + 5)$ $-7(x-5)(x+5)$

Explanation:

$- 7 x^{2} + 175$ $-7x^2+175$

Factor out the $- 7$ $-7$

= $- 7 (x^{2} - 25)$ $-7(x^2-25)$

Using the rule difference of two squares: $(x - a) (x + a) = x^{2} - a^{2}$ $(x-a)(x+a)=x^2-a^2$

= $- 7 (x - 5) (x + 5)$ $-7(x-5)(x+5)$

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How do you factor $- 7 x^{2} + 175$ $-7x^2+175$ ?

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