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 20 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!!

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