Question

For the polynomial below, 3 is a zero. H(x)=x^3-7x^2 +11x+3

Express h(x) as a product of linear factors.

Answer 1

in otherwords, factor

Answer 2

Hey, Bear: If 3 is a zero, then x-3 is a factor.

There are several ways of factoring this polynomial.

The long way is to do long division, that is, to divide (x-3) into x^3-7x^2+11x+3.
There should be no remainder to this long division if x-3 actually is one of the factors.

Another way, which I favor because it's so much faster, is synthetic division. Given that 3 is a zero, synth. div. should be a snap. Are you familiar with it?

You identify the four coefficients of the x powers in the given polynomial: {1, -7, 11, 3}.

___________
Set up synthetic division as 3 | 1 -7 11 3
___________
1

Others or I will help if you encounter any difficulty with this synthetic division. Pls try it.

Answer 3

Thanks. I'm not familiar with synthetic division but will give it a shot.

Answer 4

Try long division first if that's familiar to you.
Doing synthetic div after that will both demonstrate how you get the same numerical results with less work.
_________________________
Long div.: x-3 | x^3 - 7x^2 +11x + 3

Answer 5

x-3 is the denominator? I am watching Khan academy video in synthetic division and will see how it goes.

Answer 6

I got \[1x ^{2}-4x-1+\frac{ 0 }{x-3 }\]

Answer 7

thats correct so

h(x) = (x -3)(x^2 -4x -1)

you need to find the linear factors of the quadaratic..

which will be complex conjugates

Answer 8

use the general quadratic formula on the quadratic factor

\[x = \frac{4 \pm \sqrt{(-4)^2 - 4 \times 1 \times (-1)}}{2 \times 1}\]





simplify this and then you'll have the other 2 linear factors.

Answer 9

@campbell_st I simplified to x=\[-4\sqrt{3}\]

Answer 10

well I think you need to check it as its

\[x = 2 \pm \sqrt{5}\]

so the linear factors are

\[h(x) = (x-3)(x - (2-\sqrt{5}))(x + (2 - \sqrt{5})\]