# How do you factor by grouping 6x^4 + 7x^2 - 5?

Jun 24, 2015

$\left(2 {x}^{2} - 1\right) \left(3 {x}^{2} + 5\right)$

#### Explanation:

To factorise the expression split the middle term and rewrite the expression $6 {x}^{4} + 10 {x}^{2} - 3 {x}^{2} - 5$

$2 {x}^{2} \left(3 {x}^{2} + 5\right) - \left(3 {x}^{2} + 5\right)$

$\left(2 {x}^{2} - 1\right) \left(3 {x}^{2} + 5\right)$

Jun 25, 2015

Hi Alan, just consider ${x}^{2}$ as y, so that we get a simple quadratic expression $6 {y}^{2} + 7 y - 5$, which can be factorised easily by splitting the middle term.

#### Explanation:

Hi Alan, just consider ${x}^{2}$as y, so that we get a simple quadratic expression $6 {y}^{2} + 7 y - 5$, which can be factorised easily by splitting the middle term.

To do this we multiply the coefficient of ${y}^{2}$ and the constant term. We get -30. Now split this in two parts such that sum is +7 and the product -30. That is how we arrive at 10 and -3

Jun 28, 2015

Factor: $y = 6 {x}^{4} + 7 {x}^{2} - 5$

#### Explanation:

Call $X = {x}^{2} \to f \left(X\right) = 6 {X}^{2} + 7 X - 5$

To factor f(X) I use the new AC Method.
Converted $f ' \left(X\right) = {X}^{2} + 7 X - 30.$
Factor pairs of (-30)-> (-2, 15)(-3, 10). This sum is 7 = b.
p' = -3 and q' = 10
$p = \frac{p '}{a} = - \frac{3}{6} = - \frac{1}{2}$ and $q = \frac{q '}{a} = \frac{10}{6} = \frac{5}{3}$

Factored form: $f \left(X\right) = 6 \left(X - \frac{1}{2}\right) \left(X + \frac{5}{3}\right) = \left(2 X - 1\right) \left(3 X + 5\right)$

$f \left(x\right) = \left(2 {x}^{2} - 1\right) \left(3 {x}^{2} + 5\right) = \left(x \sqrt{2} - 1\right) \left(x \sqrt{2} + 1\right) \left(3 {x}^{2} + 5\right)$