How do you write a four term polynomial that can be factored by grouping?

2 Answers
Jul 27, 2017

Answer:

e.g.: #x^3+3x^2-4x-12 = (x-2)(x+2)(x+3)#

Explanation:

We can construct such a polynomial by reversing the factorisation.

For example, to get an example cubic, pick any numbers #a, b, c, d# and multiply out:

#(ax+b)(cx^2+d) = acx^3+bcx^2+adx+bd#

So picking #a=1#, #b=3#, #c=1#, #d=-4#, we would get:

#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)(x+2)(x+3)#

Note that in this example, I chose #c=1# and #d=-4=-2^2#, so the quadratic factor factored cleanly.

If we want clean factorisations like this, we could use the alternative pattern:

#(ax+b)(cx-d)(cx+d) = (ax+b)(c^2x^2-d^2)#

#color(white)((ax+b)(cx-d)(cx+d)) = ac^2x^3+bc^2x^2-ad^2x-bd^2#

Jul 28, 2017

Answer:

Here is one way I've done it when writing exam questions.

Explanation:

Make sure that the first two terms have a common factor. Not a constant common factor, but something involving the variable.

Warning: This may lead to a polynomial that cannot be factored completely over the integers.

For example we might start with

#14x^5-49x^3#

There is a common factor of #7x^3#.

The other factor is #2x^2-7#

For our third and fourth terms we need a multiple of #2x^2-7# In order to avoid a common factor of #x# use a constant.

We could use the following:

#5# to get #14x^5-49x^3 + 5(2x^2-7) = 14x^5-49x^3+10x^2-35#
Which factors as #(7x^3+5)(2x^2-7)#

Neither of these factor can be factored over the integers.