ALGEBRA 2 PROBLEM: polynomials?

Question

Please help me solve the following problem:

Find the polynomial whose roots are 2 more than the roots of x^4-3x^3-3x^2+31x-42=0.

Explanations on how to solvet his problem is much appreciated.

Answer 1

[Note: I worked this two different ways (and got two different answers!). I have more confidence in the first way. I thought the second way might be easier, but I no longer think so.]

This is not an easy problem. First, we need to get the four roots of the quartic. There are different ways to do it; I guessed that one or more of the roots might be easy ones, so I evaluated at a few points:

f(0) = -42
f(1) = 1 - 3 - 3 + 31 - 42 = -16
f(2) = 16 - 24 - 12 + 62 - 42 = 0

x=2 is a root, so we can reduce it to a cubic, either by synthetic division or polynomial long division. When you do that, you get

x^3 - x^2 - 5x + 21 = 0

We've already looked at x = 0, 1, and 2, so let's go the other way:

f(-1) = -1 -1 + 5 + 21 = 24
f(-2) = -8 - 4 + 10 + 21 = 19
f(-3) = -27 - 9 + 15 + 21 = 0

x = -3 is a second root. Divide that out to get a quadratic:

x^2 - 4x + 7 = 0

Complete the square:

x^2 - 4x + 4 = -7 + 4 = -3
(x-2)^2 = -3
x = 2 +/- i sqrt 3

So we have the four roots: x = 2, -3, 2 + i sqrt 3, 2 - i sqrt 3

Adding 2 to each root and forming a polynomial, we have

(x-4)(x-1)(x-4 + i sqrt 3)(x-4 - i sqrt 3)

(x^2 - 5x + 4)[(x-4)^2 - 3 i^2]

(x^2 - 5x + 4)(x^2 - 8x + 16 + 3)

(x^2 - 5x + 4)(x^2 - 8x + 19)

When you multiply these two trinomials together, you get

x^4 - 13x^3 + 63x^2 - 127x + 76 (Answer)

That's your answer, but there's another, perhaps easier, way to do it. In your original equation

x^4 - 3x^3 - 3x^2 + 31x - 42

suppose we make a substitution, y = x+2, or x = y-2. Then the equation becomes

(y-2)^4 - 3(y-2)^3 - 3(y-2)^2 + 31(y-2) - 42

(y-2)^4 - 3[(y-2)^3 + (y-2)^2] + 31(y-2) - 42

(y-2)^4 = y^4 + 4y^3 (-2) + 6y^2 (-2)^2 + 4y (-2)^3 + (-2)^4
= y^4 - 8y^3 + 24y^2 - 32y + 16

(y-2)^3 = y^3 + 3y^2 (-2) + 3y (-2)^2 + (-2)^3
= y^3 - 6y^2 + 12y - 8

(y-2)^2 = y^2 - 4y + 4

When you get all the coefficients in, and multiply it out, you get

(y^4 - 8y^3 + 24y^2 - 32y + 16) - 3(y^3 - 5y^2 + 8y - 4) + 31y - 62 - 42

y^4 - 11y^3 + 39y^2 - 25y - 76 (Another answer)

My two polynomial answers don't agree, and I'm not sure I know why. I did check the root x = 4 (or y = 4), and f(4)=0 in both cases.

My guess is that the first answer's the better one ... maybe I made an algebra mistake in the second one. Also, having worked this problem in two different ways, I think the first way is easier.

For Algebra 2, this is a tough problem.

Answer 2

Below is a link to a great explanation about how to solve this problem.

First of all, find the roots of the given equation. Then, add 2 to each one. Then find a polynomial with those roots.

Once you find an answer, you can check it using this site: http://www.quickmath.com/www02/pages/modules/equations/index.shtml

By the way, bpiguy got the right answer with the second method. The first method has a tiny error with a sign when he was finding the polynomial with the given roots.

Answer 3

x^4 - 3x^3 - 3x^2 + 31x - 42=0
Suppose the roots are α, β, γ, δ

Then sum of roots taken 1 at a time
Σα = -b/a = 3
Sum of roots taken 2 at a time
Σαβ = c/a = -3
Sum of roots taken 3 at a time
Σαβγ = -d/a = -31
Sum of roots taken 4 at a time = product of all the roots
Σαβγδ = e/a = -42

So new polynomial will have roots α + 2, β + 2, γ + 2, δ + 2

Sum of roots taken 1 at a time = α + 2 + β + 2 + γ + 2 + δ + 2
= α + β + γ + δ + 8
= Σα + 8
= 11

Sum of roots taken 2 at a time = (α + 2)(β + 2) + (α + 2)(γ + 2) + (α + 2)(δ + 2) + (β + 2)(γ + 2) + (β + 2)(δ + 2) + (γ + 2)(δ + 2)
= αβ + 2(α + β) + 4 + αγ + 2(α + γ) + 4 + αδ + 2(α + δ) + 4 + βγ + 2(β + γ) + 4 + βδ + 2(β + δ) + 4 + γδ + 2(γ + δ) + 4
= Σαβ + 6Σα + 24
= -3 + 6*3 + 24
= 39

Sum of roots taken 3 at a time = (α + 2)(β + 2)(γ + 2) + (α + 2)(β + 2)(δ + 2) + (α + 2)(γ + 2)(δ + 2) + (β + 2)(γ + 2)(δ + 2)
= αβγ + 2(αβ + αγ + βγ) + 4(α + β + γ) + 8 + αβδ + 2(αβ + αδ + βδ) + 4(α + β + δ) + 8 + αγδ + 2(αγ + αδ + γδ) + 4(α + γ + δ) + 8 + βγδ + 2(βγ + βδ + γδ) + 4(β + γ + δ) + 8
= Σαβγ + 4Σαβ + 12Σα + 32
= -31 + 4*(-3) + 12*3 + 32
= 25

Sum of roots taken 4 at a time = (α + 2)(β + 2)(γ + 2)(δ + 2)
= Σαβγδ + 2Σαβγ + 4Σαβ + 8Σα + 16
= -42 + 2*(-31) + 4*(-3) + 8*(3) + 16
= -42 - 62 - 12 + 24 + 16
= -76

New equation is x^4 - Σα * x^3 + Σαβ * x^2 - Σαβγ * x + Σαβγδ = 0
ie x^4 - 11x^3 + 39x^2 - 25x - 76 = 0

Answer 4

(x - 2)^4 - 3(x - 2)^3 - 3(x - 2)^2 + 31(x - 2) - 42 = 0

Answer 5

What do you mean to say '2 more than the roots of'.Make it clear please