# Thread: Need help with Factoring and Solving Equations #2.

1. ## Need help with Factoring and Solving Equations #2.

Hi! Ive been trying to do these problems for a while and Im stumped. I finished 13 problems and fully understand them , but I don't know how to do these problems. It would be nice if you could write each step out or the answer at the very least. Or hints or tips, like which theorem or something to use. Shortcuts.

3. Find the quadratic equation. Use only integral coefficients
-2 - (3/2)i, -1, 5

What do you do in this problem O_o

4) Solve x4 + 2x3 +14x2 + 8x + 40 = 0 , given that 2i is a root.

I think there are shortcuts for this one. But I dont know them
Is there any shortcuts?

5. Can anyone explain the Descartes Rule of Signs?

2. Descartes Rule of Sign

Example

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

x cubed has a positive sign, x squared negative, x negative, and 5 positive.

+ - - +

1 sign change from + to - and then another from - to +.

We have 2 sign changes.

Positive solutions are 2 or 0. You always have two choices from greater than or equal to 2.

(+) 2 or 0

Now make x negative.

$(-x)^3-2(-x)^2-(-x)+5=0\Rightarrow -x^3-2x^2+x+5=0$

- - + +

We have 1 sign change from - to +.

Therefore, (-) 1. We don't subtract 2 since, 1 or -1 solutions isn't possible. Only 1 is.

Since we had 2 or 0 positive, we must have 2 or 0 imaginary. This is because we know we have 1 negative and if there are 0 positive, we must have 2 imaginary.

3. Hello, donaldmax!

In problem #3, isn't that "a quartic equation"?
(And there is more than one answer.)

$\text{3. Find the quartic equation with the following roots.}$
. . . $\text{Use only integral coefficients..}$

. . . $\text{-}2 - \frac{3}{2}i,\;\text{-}1,\; 5$

It appears that there are huge gaps in your training.

If $\,a$ is a root of a function $\,f(x)$, then $(x-a)$ is a factor of $f(x).$

Complex roots occur in conjugate pairs.
. . If $a+bi$ is a root, then $a-bi$ is also a root.

Since $\text{-}2-\frac{3}{2}i$ is a root, then $\text{-}2+\frac{3}{2}i$ is also a root.

We have four roots: . $\text{-}1,\;5,\;\text{-}2-\frac{3}{2}i,\;\text{-}2 + \frac{3}{2}i$

Then we have these factors: . $\begin{Bmatrix}
x - (\text{-}1) &=& x + 1 \\ x - 5 &=& x-5\\ x - (\text{-}2-\frac{3}{2}i) &=& x + 2 + \frac{3}{2}i \\ x - (\text{-}2 + \frac{3}{2}i}) &=& x + 2 - \frac{3}{2}i \end{Bmatrix}$

Their product is: . $(x+1)(x-5)\left(x+2+\frac{3}{2}i\right)\left(x + 2 - \frac{3}{2}i\right)$

. . . . . . . . . . $=\;x^4 - \frac{59}{4}x^2 - 45x - \frac{125}{4}$

We want integral coefficients; multiply by 4.

. . $f(x) \;=\;4x^4 - 59x^2 - 180x - 125$

$\text{4) Solve: }\: x^4 + 2x^3 +14x^2 + 8x + 40 \:=\: 0$
. . . $\text{given that }2i\text{ is a root.}$

Since $2i$ is a root, then $-2i$ is also a root.

Then $(x-2i)$ and $(x + 2i)$ are factors.
. . That is, $(x-2i)(x+2i) \:=\:x^2+4$ is a factor.

Using long division, the quartic factors into: . $(x^2+4)(x^2+2x+10)$

. . $x \:=\:\dfrac{\text{-}2 \pm\sqrt{4-40}}{2} \:=\:\dfrac{\text{-}2\pm6i}{2} \:=\:\text{-}1 \pm 3i$
And we have the four root of the quartic: . $\pm2i,\;\text{-}1 \pm3i$