Solve x^4-3x^3-8x^2+12x+16 = 0 given roots x = -1, 4

[Jodene222]

Solve each equation given the 2 roots

-1, 4

x^4 -3x^3 - 8x^2 + 12x + 16 = 0

using -1 , I get:

3x^3 + 3x^2 - 9x + 3

now I need to use the quadratic equation.
so a=2 b=9 c=3
i do not get the correct answers.

the correct answers are 1, -2/3, -1 times square root of 2 (sorry, not sure how to type square root sign)

 

[stapel]

The Quadratic Formula (not "Equation") applies only to quadratics, not to cubics, so I'm not clear on what you applied, or to what...?

I will guess that you used synthetic division (with x = -1) or long polynomial division (with x + 1) to reduce the quartic to a cubic. It is unfortunate that you did not show (or describe) your work, since this prevents us from being able to find your error(s). The synthetic division should have looked like this:

. . . . .\(\displaystyle \begin{array}{r|rrrrr}-1&1&-3&-8&12&16\\ & & -1&4&4&-16\\\hline &1&-4&-4&16&0 \end{array}\)
...at which point, you should have used the other root they gave you. Only after you'd reduced the cubic to a quadratic could the Quadratic Formula be brought to bear.

Please reply with clarification. Thank you!

Eliz.

 

[soroban]

Re: need help

Hello, Jodene!

Solve the equation, given the 2 roots: .\(\displaystyle x \;=\;-1,\:4\)

. . \(\displaystyle x^4 -3x^3 - 8x^2 + 12x + 16\:= \:0\)

Using -1 , I get: .\(\displaystyle 3x^3 + 3x^2 - 9x + 3\) . . . . How?

now I need to use the quadratic equation. . . . . on a cubic equation?

so: a = 2, b = 9, c = 3 . . . . Why does a = 2 ?

i do not get the correct answers. . . . . Of course not.

The correct answers are 1, -2/3, -1 times square root of 2 . . . . no, they are not

Let's start again . . .


\(\displaystyle \text{Since }x = -1\text{ is a root, then: }\x + 1)\text{ is a factor.}\)

. . \(\displaystyle \text{We have: }\;x^4 - 3x^3 - 8x^2 + 12x + 16 \;=\;(x+1)(x^3-4x^2-4x+16)\)


\(\displaystyle \text{Since }x = 4\text{ is a root, then: }\x - 4)\text{ is a factor.}\)

. . \(\displaystyle \text{Hence: }\;x^4-3x^3 - 8x^2 + 12x + 16 \;=\;(x+1)(x-4)(x^2-4)\)


\(\displaystyle \text{The equation becomes: }\;(x+1)(x-4)(x-2)(x+2) \:=\:0\)

. . \(\displaystyle \text{Therefore, the roots are: }\:\boxed{x \;=\;-1,\:4,\:2, \:-2}\)

 

[Jodene222]

Let me clarify: following is the corrected equation.

Solve each equation given the two indicated roots.

3x^4 - 5x^3 -7x^2 - 3x + 2 roots are 1 , -2/3

I know I need to find the depressed equation by using synthetic division.
when i did this i got the following depressed equation when x = 1

3x^3 + 8x^2 + x - 2 = 0
now I need to use the quadratic formula but not sure how to do this with a cube.

 

[tkhunny]

now I need to use the quadratic formula but not sure how to do this with a cube.

This makes so little sense. You DON'T use the QUADRATIC formula with a CUBIC. You must have a QUADRATIC.

You have used x = 1. Now use x = -2/3 and depress it again. This will give you the desired QUADRATIC.

 

[stapel]

Let me clarify: following is the corrected equation.

Solve each equation given the two indicated roots.

3x^4 - 5x^3 -7x^2 - 3x + 2 roots are 1 , -2/3

Um... This doesn't look like a "corrected" equation; this appears to be a different exercise...? :shock:

It seems fairly clear that you haven't read the explanations provided earlier (else you wouldn't have taken the exact same wrong steps), so I'm guessing you're not interested in that part...? So maybe Soroban will give you the answer for this one, too. :roll:

But if he doesn't, or if you're interested in learning how to do this on your own (in case your tests aren't online, so you can't come here during the test to get the answers), then please study the explanations provided (rather than just the solution) and kindly reply, clearly showing your work and reasoning. :idea:

Thank you!

Eliz.