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

3 Answers
Jun 24, 2015

Answer:

#(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)#

Jun 25, 2015

Answer:

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

Jun 28, 2015

Answer:

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)#