How do you factor #3y^4-11y^2-42# completely?

1 Answer
George C.
Oct 19, 2016

#3y^4-11y^2-42 = (3y^2+7)(y^2-6)#

#color(white)(3y^4-11y^2-42) = (3y^2+7)(y-sqrt(6))(y+sqrt(6))#

#color(white)(3y^4-11y^2-42) = (sqrt(3)y-sqrt(7)i)(sqrt(3)y+sqrt(7)i)(y-sqrt(6))(y+sqrt(6))#

Explanation:

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

We use this once or twice below...

Given:

#3y^4-11y^2-42#

Start by treating as a quadratic in #y^2# and use an AC method to factor it.

Look for a pair of factors of #AC = 3*42 = 126# which differ by #B=11#.

The pair #18, 7# works.

Use this pair to split the middle term and factor by grouping:

#3y^4-11y^2-42 = 3y^4-18y^2+7y^2-42#

#color(white)(3y^4-11y^2-42) = (3y^4-18y^2)+(7y^2-42)#

#color(white)(3y^4-11y^2-42) = 3y^2(y^2-6)+7(y^2-6)#

#color(white)(3y^4-11y^2-42) = (3y^2+7)(y^2-6)#

#color(white)(3y^4-11y^2-42) = (3y^2+7)(y^2-(sqrt(6))^2)#

#color(white)(3y^4-11y^2-42) = (3y^2+7)(y-sqrt(6))(y+sqrt(6))#

The remaining quadratic has no linear factors with Real coefficients.

If you really want to factor it, you can use Complex coefficients as follows:

#3y^2+7 = (sqrt(3)y)^2 - (sqrt(7)i)^2#

#color(white)(3y^2+7) = (sqrt(3)y-sqrt(7)i)(sqrt(3)y+sqrt(7)i)#

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