# Thread: Yet another polynomial problem!! sigh...

1. ## Yet another polynomial problem!! sigh...

Yea i have another polynomial problem..I tried it 3 times sigh....
Solve the equation $x^{4}+2x^{3}-21x^{2}-22x+40=0$ given that the roots are in an arithmetic progression. Let the roots be a-3d, a-d, a+d, a+3d

Thanks a lot

2. Use Cardano Vieta relations.

Fernando Revilla

3. Hi, um I actually googled cardano Vieta relations and I think that is outside the scope of my high school course, this is a Mathematics Extension 2 question (for those who are familiar with the NSW (Australia) maths high-school course). Anyway thanks

4. Originally Posted by aonin
Yea i have another polynomial problem..I tried it 3 times sigh....
Solve the equation $x^{4}+2x^{3}-21x^{2}-22x+40=0$ given that the roots are in an arithmetic progression. Let the roots be a-3d, a-d, a+d, a+3d

Thanks a lot
You know that $\displaystyle a-3d, a-d, a+d, a+3d$ are all roots of the polynomial $\displaystyle P(x) = x^4 + 2x^3 - 21x^2 - 22x + 40$.

So $\displaystyle P(a-3d) = 0, P(a - d) = 0, P(a + d) = 0, P(a + 3d) = 0$.

Substitute the values and you will end up with a system of equations to solve for $\displaystyle a$ and $\displaystyle d$.

5. lol is it that easy? nvm...i stuck by trying to solve it using sums and products relations between roots and coefficients. But I guess what you suggested would solve it except power of 4 and 3 are very tedious.

6. Originally Posted by aonin
lol is it that easy? nvm...i stuck by trying to solve it using sums and products relations between roots and coefficients. But I guess what you suggested would solve it except power of 4 and 3 are very tedious.
Those are the relations I was talking about:

$x_1+x_2+x_3+x_4=4a=\dfrac{-2}{1}=-2$

so, $a=-1/2$

Could you continue?

Fernando Revilla

7. Hello, aonin!

$\text{Solve the equation: }\:x^4+2x^3-21x^2-22x+40=0$
$\text{ given that the roots are in an arithmetic progression.}$

$\text{Let the roots be: }\: a-3d,\;a-d,\;a+d,\;a+3d$ . . . . but why?

The terms would be: . $a-2d,\;a-d,\;a+d,\;a+2d$

Why force the A.P. to go up by twice some number?

And do they really expect us to plug those four expressions into the polynomial?

Using the Factor Theorem: . $(x+5)(x+2)(x-1)(x-4) \:=\:0$

Therefore, the roots are: . $x \;=\;\text{-}5,\:\text{-}2,\:1,\:4$

8. Originally Posted by Soroban
Hello, aonin!

The terms would be: . $a-2d,\;a-d,\;a+d,\;a+2d$

Why force the A.P. to go up by twice some number?

And do they really expect us to plug those four expressions into the polynomial?

Using the Factor Theorem: . $(x+5)(x+2)(x-1)(x-4) \:=\:0$

Therefore, the roots are: . $x \;=\;\text{-}5,\:\text{-}2,\:1,\:4$

I expect that's what they were told to do in the problem...

9. Certainly the equation can be easily solved using Ruffini's rule. However the sense of the problem is surely to use the fact that the roots are in arithmetic progression. An this, in order to see if the student knows the relations between roots an coefficients. In this case the relations:

$\begin{Bmatrix} x_1+x_2+x_3+x_4=-2\\x_1x_2x_3x_4=40\end{matrix}$

lead to $a=-1/2$ and to the biquadratic equation:

$\left(\dfrac{1}{4}-9d^2\right)\left(\dfrac{1}{4}-d^2\right)=40$

Of course it is not necessary to write the difference of the progression as $2d$ , but it is correct.

Fernando Revilla