Solving a cubic function by
factoring: using the sum or
difference of two cubes.
By Diane Webb
What is a cube?
27
1
64
8
125
Factoring the sum or difference
of two cubes:
(a³+b³) a³ is a perfect cube since a*a*a = a³
b³ is a perfect cube since b*b*b = b³
Always remember that the factors of the sum or
difference of two cubes is always a (binomial) and a
(trinomial).
(a³+b³)=(binomial)(trinomial)
To find the two factors,
let’s do the following:
First the binomial: Take the cubed root of each monomial
within the problem.
l
Since, the cubed root of a³ = a and the cubed root of
b³ = b.
(a³+b³)=(a+b)(trinomial)
Now, the trinomial.
The first term of the trinomial is
the first term of the binomial squared.
The first term of the
binomial is “a” and a*a = a²
The second term of the trinomial is the opposite
of the product of the two terms of the binomial.
The product of “a” and “b” is “ab” and
(a³+b³)=(a+b)(a²-ab+__)
then the opposite tells you to
change the sign of the product.
The third term of the polynomial is the 2nd term
of the binomial squared.
(a³+b³)=(a+b)(a²+__+__)
(a³+b³)=(a+b)(a²-ab+b²)
The second term of the binomial is
“b” and b*b=b²
Factor (x³-8)
Binomial factor is (x-2)
Trinomial factor is (x²+2x+4)
Remember that the trinomial is not factorable.
Factored form: (x³-8)=(x-2)(x²+2x+4)
Is it possible to check our answers?
Remember that you may check using either the Remainder
Theorem or division.
Remainder theorem: If P(x)=x³-8 and the factor is (x-2),
Then P(2)=(2)³-8
=8–8
=0
Since the remainder is 0, then x-2 is a factor.
Synthetic division:
2 1 0 0 -8
2 4 8
1 2 4 0
This tells you two things: 1. (X-2) is a factor since the
remainder is 0. 2. The quotient is x²+2x+4 which is
the trinomial factor of the cubic polynomial.
What about 2x³+2?
First of all, you can not forget that the
GCF must be factored out of the cubic
function. SO, what is the GCF of
2x³ and 2?
2 is the GCF. Now, factor
the two first: 2(x³+1)
Look at the binomial. Is it a difference
or the sum of two cubes. If yes, factor the
expression.
2x³+2 = 2(x³+1) = 2(x+1)(x²-x+1)
Now that we can factor the
sum and difference of two
cubes, let us solve them.
Remember to solve a cubic equation, we
need to use our factors. Set your
factors equal to 0. Then solve for x.
• Go back to the the previous problem:
• 2x³+2=0
•
•
•
•
•
•
Factored form: 2(x+1)(x²-x+1)=0
Set the factors equal to 0.
2=0 (x+1)=0 (x²-x+1)=0
20 so it is not part of our solution.
X+1=0 so x = -1.
What about x²-x+1=0? Remember we talked
about the fact that it is not factorable. How
do we solve that quadratic?
• QUADRATIC FORMULA!!