How do I factor a polynomial completely through grouping?

1 Answer
Mar 8, 2017

Answer:

Here are a couple of examples explained...

Explanation:

This is often easier for cubics than quadratics.

#color(white)()#
Cubic

For example, given:

#x^3+3x^2-4x-12#

You might notice that the ratio of the first and second terms is the same as that between the third and fourth terms.

So we can group the terms in pairs to find a factorisation:

#x^3+3x^2-4x-12 = (x^3+3x^2)-(4x+12)#

#color(white)(x^3+3x^2-4x-12) = x^2(x+3)-4(x+3)#

#color(white)(x^3+3x^2-4x-12) = (x^2-4)(x+3)#

#color(white)(x^3+3x^2-4x-12) = (x^2-2^2)(x+3)#

#color(white)(x^3+3x^2-4x-12) = (x-2)(x+2)(x+3)#

Note the steps:

  • Split the quadrinomial into two bracketed binomials.

  • For each binomial, separate out the common factor of the terms.

  • Combine the expressions to give a factorisation as a quadratic binomial multiplied by a linear one.

  • Factor the quadratic binomial as a difference of squares.

Note that the last step may not be possible using real numbers, for example #(x^2+4)# would not factorise any further without complex coefficients.

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Quadratic

To factor a quadratic by grouping you typically need to work out how to split the middle term.

For example, given:

#42x^2+41x-105#

we can use an AC method:

Look for a pair of factors of #AC = 42*105 = 4410# which differ by #B=41#.

The pair #90, 49# works, in that #90*49 = 4410# and #90-49 = 41#.

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

#42x^2+41x-105 = 42x^2+90x-49x-105#

#color(white)(42x^2+41x-105) = (42x^2+90x)-(49x+105)#

#color(white)(42x^2+41x-105) = 6x(7x+15)-7(7x+15)#

#color(white)(42x^2+41x-105) = (6x-7)(7x+15)#