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Polynomials

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Find a, b and the other factor

Guest Nov 8, 2017
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#1
+82634
+2

Since the first term is x^3  and the last term  is  -8....

The other factor must be (x - 4)

So....let's expand this

(x + 1) (x + 2) (x - 4)  =   x^3 - x^2 - 10 x - 8

So  a  = 1    and b  =  10

CPhill  Nov 8, 2017
edited by CPhill  Nov 8, 2017
edited by CPhill  Nov 8, 2017
#2
+18936
+2

Polynomials

Find a, b and the other factor

$$\text{factor } (x+1) \Rightarrow \text{let } x_1 = -1 \\ \text{factor } (x+2) \Rightarrow \text{let } x_2 = -2$$

Vieta's formulas:

$$\begin{array}{|rcll|} \hline -8 &=& -(x_1x_2x_3) \\ 8 &=& x_1x_2x_3 \\ 8 &=& (-1)(-2)x_3 \\ 8 &=& 2x_3 \\ \mathbf{x_3} &\mathbf{=}& \mathbf{4} \\ x_3 = 4 \Rightarrow \text{factor } (x-4) \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline -a &=& -(x_1+x_2+x_3) \\ &=& (-1)+(-2)+4 \\ &=& -3+4 \\ \mathbf{a} &\mathbf{=}& \mathbf{1} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline -b &=& x_1x_2+x_1x_3+x_2x_3 \\ b &=& -(x_1x_2+x_1x_3+x_2x_3) \\ &=& -\Big((-1)(-2)+(-1)4+(-2)4)\Big) \\ &=& - (2-4-8) \\ &=& - (2-12) \\ &=& - (-10) \\ \mathbf{b} &\mathbf{=}& \mathbf{10} \\ \hline \end{array}$$

heureka  Nov 8, 2017
#3
+82634
+2

Thanks, heureka....here's one other way

Let the polynomial  be   a(x + 1)(x + 2)(x - q)

Where q is the unknown root

Since the lead coefficient is 1, a = 1

So. expanding this, we have

(x^2 + 3x + 2)(x - q )  =   x^3 + 3x^2 + 2x - qx^2 - 3qx - 2q

Equating coefficients, we have

(3 - q) = -a

(2 -3q)  = -b

-2q  = -8   ⇒   q  = 4   = the unknown root

So

(3 - 4)  =  -a   ⇒  -1  = -a  ⇒  1  =  a

( 2 - 3(2) )  =  -b  ⇒  -10  = -b  ⇒  10 =  b

CPhill  Nov 8, 2017

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