How do you write a four term polynomial that can be factored by grouping?
2 Answers
e.g.:
Explanation:
We can construct such a polynomial by reversing the factorisation.
For example, to get an example cubic, pick any numbers
(ax+b)(cx2+d)=acx3+bcx2+adx+bd
So picking
x3+3x2−4x−12=(x3+3x2)−(4x+12)
x3+3x2−4x−12=x2(x+3)−4(x+3)
x3+3x2−4x−12=(x2−4)(x+3)
x3+3x2−4x−12=(x−2)(x+2)(x+3)
Note that in this example, I chose
If we want clean factorisations like this, we could use the alternative pattern:
(ax+b)(cx−d)(cx+d)=(ax+b)(c2x2−d2)
(ax+b)(cx−d)(cx+d)=ac2x3+bc2x2−ad2x−bd2
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
There is a common factor of
The other factor is
For our third and fourth terms we need a multiple of
We could use the following:
Which factors as
Neither of these factor can be factored over the integers.
