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

What is a cube? 27 1 125 8 64

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)

let’s do the following: 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 (a³+b³)=(a+b)(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. (a³+b³)=(a+b)(a²-ab+__) The third term of the polynomial is the 2nd term of the binomial squared. (a³+b³)=(a+b)(a²-ab+b²) The second term of the binomial is “b” and b*b=b²

Factor (x³-8) Factored form: (x³-8)=(x-2)(x²+2x+4) 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!!