Synthetic Division & Factoring

For x − 4 to be a factor of the given polynomial, then I must have x = 4 as a zero. (Remember that this is how we solved quadratics by factoring: We'd find the two factors, set each of the factors equal to zero, and solve. Here, we're working backwards from zeroes to factors.)

Using this information, I'll do the synthetic division with x = 4 as the test zero on the left:

Since the remainder is zero, then x = 4 is indeed a zero of −2x⁵ + 6x⁴ + 10x³ − 6x² − 9x + 4, so:

Remember that, if x = b is a zero, then x − b is a factor. I'll start by using the Rational Roots Test (and maybe a quick graph on my calculator) to find a good value to test as possibly being a zero (that is, possibly an x-intercept on the graph). I'll try x = 1:

This division gives a zero remainder, so x = 1 must be a zero, which means that x − 1 is a factor. Since I divided a linear factor (namely, x − 1) out of the original polynomial, then my result has to be a cubic:

So I need to find another zero before I can apply the Quadratic Formula. I'll try x = −2:

Since I got a zero remainder, then x = −2 is a zero, so x + 2 is a factor. Plus, I'm now down to a quadratic, 15x² − 14x − 8, which happens to factor as:

Then the fully-factored form of the original polynomial (namely, 15x⁴ + x³ − 52x² + 20x + 16) is:

Since they have given me one of the zeroes, I'll use synthetic division to divide it out:

(You will probably want to use scratch paper for the computations required when manipulating the radical root.)

I only get these square-root answers from using the Quadratic Formula. Since the square-root part of the Formula is preceded by a "plus-minus" sign, then these square-root answers always come in pairs. Thus, if is a root, then so also must be a root.

So my next step is to divide by :

I had started with a fourth-power polynomial. After the first division, I was left with a cubic (with very nasty coefficients!). After the second division, I'm now down to a quadratic (x² + 0x − 5, or just x² − 5), which I know how to solve:

Then the full solution is:

They have given me a zero, so I'll use synthetic division and divide out 2 − i:

(You will probably want to use scratch paper for the computations required when doing complex division.)

What should I try next? I'll use the fact that, to arrive at a zero of 2 − i, the author of this exercise must have used the Quadratic Formula at some point, and the Quadratic Formula always spits out complex-valued answers in pairs. That is, you get the imaginary part (the part with the i in it) from having a negative inside the "plus or minus square-root of" part of the Formula. This means that, since 2 − i is a zero, then 2 + i must also be a zero. So I'll divide by 2 + i:

This leaves me with a cubic, so I'll need to find another zero on my own. (That is, I can't apply the Quadratic Formula yet.) I can use the Rational Roots Test to generate a list of potential zeroes, and a quick graph of y = x³ − 2x² − 2x + 1 can help me narrow my search.

After checking the graph in my calculator, I decide that I will try x = −1:

Now I'm down to a quadratic (namely, x² −  3x + 1) which happens not to factor, so instead I'll apply the Quadratic Formula to get:

Then all the zeroes of x⁵ − 6x⁴ + 11x³ − x² − 14x + 5 are given by: