How do you factor $5 x^{4} + 2 x^{2} - 7$ $5x^4 + 2x^2 - 7$ ?

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How do you factor $5 x^{4} + 2 x^{2} - 7$ $5x^4 + 2x^2 - 7$ ?

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Answer 1 · 2018-08-11T19:00:06

$5 x^{4} + 2 x^{2} - 7 = (x + 1) (x - 1) (5 x^{2} + 7)$ $5x^4+2x^2-7=(x+1)(x-1)(5x^2+7)$

Explanation:

$5 x^{4} + 2 x^{2} - 7 = 5 x^{4} + 7 x^{2} - 5 x^{2} - 7$ $5x^4+2x^2-7=5x^4+7x^2-5x^2-7$
because,

$7 \times - 5 = - 35$ $7 xx -5 = -35$

$7 + (- 5) = 2$ $7 + (-5) = 2$

$= x^{2} (5 x^{2} + 7) - 1 \times (5 x^{2} + 7)$ $=x^2(5x^2+7)-1xx(5x^2+7)$

$= (x^{2} - 1) \times (5 x^{2} + 7)$ $=(x^2-1)xx(5x^2+7)$

Further,

$x^{2} - 1 = (x + 1) (x - 1)$ $x^2-1=(x+1)(x-1)$

Thus,

$5 x^{4} + 2 x^{2} - 7 = (x + 1) (x - 1) (5 x^{2} + 7)$ $5x^4+2x^2-7=(x+1)(x-1)(5x^2+7)$

Answer 2 · 2018-08-11T19:30:01

$(x - 1) (x + 1) (5 x^{2} + 7)$ $(x-1)(x+1)(5x^2+7)$

Explanation:

$let x^{2} = u$ $"let "x^2=u$

$= 5 u^{2} + 2 u - 7$ $=5u^2+2u-7$

$factor the quadratic using the a-c method$ $"factor the quadratic using the a-c method"$

$the factors of the product 5 \times - 7 = - 35$ $"the factors of the product "5xx-7=-35$

$which sum to + 2 are - 5 and + 7$ $"which sum to "+2" are "-5" and "+7$

$use these factors to split the middle term$ $"use these factors to split the middle term"$

$5 u^{2} - 5 u + 7 u - 7 \leftarrow factor by grouping$ $5u^2-5u+7u-7larrcolor(blue)"factor by grouping"$

$= 5 u (u - 1) + 7 (u - 1)$ $=color(red)(5u)(u-1)color(red)(+7)(u-1)$

$take out the common factor (u - 1)$ $"take out the "color(blue)"common factor "(u-1)$

$= (u - 1) (5 u + 7)$ $=(u-1)(color(red)(5u+7))$

$change u back to x^{2}$ $"change u back to "x^2$

$= (x^{2} - 1) (5 x^{2} + 7)$ $=(x^2-1)(5x^2+7)$

$(x^{2} - 1) is a difference of squares$ $(x^2-1)" is a "color(blue)"difference of squares"$

$= (x - 1) (x + 1) (5 x^{2} + 7)$ $=(x-1)(x+1)(5x^2+7)$

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How do you factor $5 x^{4} + 2 x^{2} - 7$ $5x^4 + 2x^2 - 7$ ?

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