Linear factors?

Question

For the polynomial below, 3 is a zero.

h(x)=x^3-9x^2+25x-21
Express h(x) as a product of linear factors.

Answer 1

Perform synthetic division using 3 in the "notch" of that process.  This will develop the depressed polynomial factor, a quadratic, from which you can find the other two roots.

 

Then the answer to this question will be h(x) = (x-root1)(x-root2)(x-root3) = 0; just fill in the specific numbers for the roots.

Answer 2

If 3 is a root of f(x), then x-3 is a factor of f(x).  Use synthetic division to divide f(x) by x-3.

 

3⌋   1     -9     25    -21

   ______3___-18___21

      1     -6      7      0

 

The last number in the bottom row is the remainder when f(x) is divided by x-3.  Since the remainder is zero, x-3 is a factor of f(x).

 

The first three numbers in the bottom row are the coefficients of the quotient when f(x) is divided by x-3.

 

Since f(x) has degree 3 and we divided by a first degree polynomial, the quotient has degree 2.  So, the quotient is x² - 6x + 7.

 

Therefore, f(x) = (x-3)(x²-6x+7)

 

Using the quadratic formula, the zeros of x²-6x+7 = 0 are:

        x = [6 ± √8]/2 = [6 ± 2√2]/2 = 3 ± √2

 

So, f(x) = (x - 3)[x - (3+√2)][x - (3-√2)]