please hurry its due in 20 minutes Prove that a cubic equation x3 + ax2 + bx+ c = 0 has 3 roots by finding the roots. Let x = r be a root of the cubic. Factor out x - r using polynomial long division. Then use the quadratic equation.

Yes, x – 4 is a factor of –2x5 + 6x4 + 10x3 – 6x2 – 9x + 4

Find all the factors of 15x4 + x3 – 52x2 + 20x + 16 by using synthetic division.

Remember that, if x = a is a zero, then x – 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  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: 15x3 + 16x2 – 36x – 16. So I need to find another zero before I can apply the Quadratic Formula. I'll try x = –2:

completed division

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:

(3x – 4)(5x + 2)

Then the fully-factored form of the original polynomial is:

15x4 + x3 – 52x2 + 20x + 16

= (x – 1)(x + 2)(3x – 4)(5x + 2)

Given that  x = -3 + sqrt(11)   is a zero of x4 + 6x3 – 7x2 – 30x + 10, fully solve the

equation x4 + 6x3 – 7x2 – 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 © 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 – 5, or just x2 – 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 – i is a zero of x5 – 6x4 + 11x3 – x2 – 14x + 5, fully solve the

equation  x5 – 6x4 + 11x3 – x2 – 14x + 5 = 0.

They have given us a zero, so I'll use synthetic division and divide out 2 – 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 – 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 – i is a zero, then 2 + i must also be a zero.  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 – 2x2 – 2x + 1 can help. I will try x = –1:

completed division

Now I'm down to a quadratic (x2

Step-by-step explanation:

3< x< 5

step-by-step explanation:

|x - 4| < 1

there are two solutions, one positive and one negative.   remember to flip the inequality when taking the negative solution

x-4 < 1       x-4 > -1

add 4 to each side

x-4+4 < 1+4       x-4+4 > -1+4

x< 5                 x> 3

3< x< 5

the cat is blue because the probability of extension towards the prefix of the diameter of the circle could only mathematically equal to the sum of the roots of the carrots from   your garden.

step-by-step explanation: