x = 5
x = 2
x = -6
Step-by-step explanation:
Use synthetic division to determine whether x β 4 is a factor of:
β2x5 + 6x4 + 10x3 β 6x2 β 9x + 4
For x β 4 to be a factor, you must have x = 4 as a zero. Using this information, I'll do the synthetic division with x = 4 as the test zero on the left:
completed division
Since the remainder is zero, then x = 4 is indeed a zero of β2x5 + 6x4 + 10x3 β 6x2 β 9x + 4, so:
Yes, x β 4 is a factor of β2x5 + 6x4 + 10x3 β 6x2 β 9x + 4
Find all the factors of 15x4 + x3 β 52x2 + 20x + 16 by using synthetic division.
Remember that, if x = a is a zero, then x β a is a factor. So use the Rational Roots Test (and maybe a quick graph) to find a good value to test for a zero (x-intercept). I'll try x = 1:
completed division
This division gives a zero remainder, so x = 1 must be a zero, which means that Β x β 1 is a factor. Since I divided a linear factor (namely, x β 1) out of the original polynomial, then my result has to be a cubic: 15x3 + 16x2 β 36x β 16. So I need to find another zero before I can apply the Quadratic Formula. I'll try x = β2:
completed division
Since I got a zero remainder, then x = β2 is a zero, so x + 2 is a factor. Plus, I'm now down to a quadratic, 15x2 β 14x β 8, which happens to factor as:
(3x β 4)(5x + 2)
Then the fully-factored form of the original polynomial is:
15x4 + x3 β 52x2 + 20x + 16
= (x β 1)(x + 2)(3x β 4)(5x + 2)
Given that Β x = -3 + sqrt(11) Β is a zero of x4 + 6x3 β 7x2 β 30x + 10, fully solve the
equation x4 + 6x3 β 7x2 β 30x + 10 = 0.
Since they have given me one of the zeroes, I'll use synthetic division to divide it out:
completed division
(You will probably want to use scratch paper for the computations required when manipulating the radical root.) Copyright Β© Elizabeth Stapel 2002-2011 All Rights Reserved
Since you only get these square-root answers by using the Quadratic Formula, and since the square-root part of the Formula is preceded by a "plus-minus" sign, then these square-root answers must always come in pairs. Thus, if x = -3 + sqrt(11) is a root, then so also must x = -3 - sqrt(11) be a root. So my next step is to divide by x = -3 - sqrt(11):
completed division
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 β 5, or just x2 β 5), which I know how to solve:
x = +/- sqrt(5)
Then the full solution is:
x = -3 +/- sqrt(11), +/- sqrt(5)
If you have studied complex numbers, then you may see a problem of the following type.
Given that 2 β i is a zero of x5 β 6x4 + 11x3 β x2 β 14x + 5, fully solve the
equation Β x5 β 6x4 + 11x3 β x2 β 14x + 5 = 0.
They have given us a zero, so I'll use synthetic division and divide out 2 β i:
completed division
(You will probably want to use scratch paper for the computations required when doing complex division.)
Recall that, to arrive at a zero of 2 β i, they must have used the Quadratic Formula, which always spits out complex answers in pairs. That is, you get the imaginary part (the part with the "i") from having a negative inside the "plus or minus square-root of" part of the Formula. This means that, since 2 β i is a zero, then 2 + i must also be a zero. Β So I'll divide by 2 + i:
completed division
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 help find potential zeroes, and a quick graph of x3 β 2x2 β 2x + 1 can help. I will try x = β1:
completed division
Now I'm down to a quadratic (x2 β 3x + 1, which happens not to factor), so I'll apply the Quadratic Formula to get:
x = (3 +/- sqrt(5))/2
Then all the zeroes of x5 β 6x4 + 11x3 β x2 β 14x + 5 are given by:
x = 2 - i, 2 + i, (3 - sqrt(5))/2, (3 + sqrt(5))/2, -1
