Solving a cubic function by factoring: using the sum or difference of two cubes.

1
Solving a cubic function by factoring using the
sum or difference of two cubes.

• By Diane Webb

2

What is a cube?
27
1
125
8
64
3
Factoring the sum or difference of two cubes
(a³b³)
a³ is a perfect cube since aaa a³
b³ is a perfect cube since bbb b³
Always remember that the factors of the sum or
difference of two cubes is always a (binomial)
and a (trinomial).
(a³b³)(binomial)(trinomial)
4
To find the two factors, lets 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.
(a³b³)(ab)(trinomial)
5
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 aa a²
(a³b³)(ab)(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 then the
opposite tells you to change the sign of the
product.
(a³b³)(ab)(a²-ab__)
The third term of the polynomial is the 2nd
term of the binomial squared.
(a³b³)(ab)(a²-abb²)
The second term of the binomial is b and bbb²
6
Factor (x³-8)

• Binomial factor is (x-2)

Trinomial factor is (x²2x4)

• Remember that the trinomial is not factorable.

Factored form (x³-8)(x-2)(x²2x4)
7
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 x²2x4
which is the trinomial factor of the cubic
polynomial.
8
What about 2x³2?
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(x³1)
Look at the binomial. Is it a difference or the
sum of two cubes. If yes, factor the expression.
2x³2 2(x³1) 2(x1)(x²-x1)
9
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.
10

• Go back to the the previous problem
• 2x³20
• Factored form 2(x1)(x²-x1)0
• Set the factors equal to 0.
• 20 (x1)0 (x²-x1)0
• 2?0 so it is not part of our solution.
• X10 so x -1.
• What about x²-x10? Remember we talked
• about the fact that it is not factorable. How
  do we solve that quadratic?

11

• QUADRATIC FORMULA!!