Question

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

Answer 1

`(2x^2 -1)(3x^2 +5)`

Explanation:

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

`2x^2( 3x^2 +5) - (3x^2 +5)`

`(2x^2 -1)(3x^2 +5)`

Answer 2

Hi Alan, just consider `x^2` as y, so that we get a simple quadratic expression `6y^2 +7y-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 `6y^2 +7y-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

Answer 3

Factor: `y = 6x^4 + 7x^2 - 5`

Explanation:

Call `X = x^2 -> f(X) = 6X^2 + 7X - 5`

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

Factored form: `f(X) = 6(X - 1/2)(X + 5/3) = (2X - 1)(3X + 5)`

`f(x) = (2x^2 - 1)(3x^2 + 5) = (xsqrt2 - 1)(xsqrt2 + 1)(3x^2 + 5)`