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Question

A. Timmy solved the polynomial equation x⁵ - x⁴ - 2x³ + 2x² - 24x + 24 = 0 and his solutions were . Explain his error(s) and list the correct solutions below.

B. Apply the function transformation of g(x) = f(x - 2) + 4 to f(x) = x³ - 8. Describe the transformations and find the final form of the function after the transformation.

Answer 1

I wasn't sure if the following comment should appear first or after showing how to factor the polynomial, so I leave it first.

When all of the coefficients of the polynomial are real integers, an irrational solution must have its conjugate, and an imaginary solution must have its conjugate, so we also have solutions -√6 and -2i.

As each of the other solutions tells you, the other root is 1, not -1

Thus, our roots are 1, √6, 2i, -√6 and -2i.

Note that the product of our roots is -24, which is the negative of the constant coefficient divided by the leading coefficient (we use the negative for odd leading powers).

The easiest way to factor the polynomial would be to notice how each pair of roots has the negative and positive of the same coefficient.

Thus, (x - 1) will be a factor.

Then,

x⁵ - x⁴ - 2x³ + 2x² - 24x + 24 = (x - 1)(x^4 - 2x^2 - 24) = (x-1)(x^2 - 6)(x^2 + 4) =

(x - 1)(x - √6)(x + √6)(x - 2i)(x + 2i)

If f(x) = x^3 - 8, f(x-2) + 4 = (x-2)^3 - 8 + 4 = x^3 - 6 x^2 + 12 x - 8 - 8 + 4 = x^3 - 6 x^2 + 12 x - 12

Answer 2

The roots are 1, √6, -√6

If you divide the original equation to the expression of (x-1)(x²-6) which is equal to x³-x²-6x+6, then the results would be :

x⁵-x⁴-2x³+2x²-24x+24 ÷ x³-x²-6x+6= x²+4

Thus, (x²+4)(x-1)(x²-6)=0

And the last roots would be x²+4=0

X²+4=0, then x=±2i

Answer 3

1 is a root, not -1

±√6 are both roots (quadratic surd roots occur in conjugate pairs

complex roots occur in conjugate pairs

x³-6x²+12x-12