My Math Forum Solving Cubic Equations

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 April 17th, 2011, 04:24 PM #1 Newbie   Joined: Mar 2011 Posts: 12 Thanks: 0 Solving Cubic Equations Consider the polynomial p(x)=x³+ax²+bx-12 Given that (x+3) and (x-4) are factors of p(x), factorise p(x) completely. I'm not sure whether I should sub in -3 and 4 or whether I should divide the original polynomial by x+3 and x-4.
April 17th, 2011, 05:09 PM   #2
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Re: Solving Cubic Equations

Hello, chris99191!

Quote:
 $\text{Consider the polynomial: }\:p(x)\:=\:x^3\,+\,ax^2\,+\,bx\,-\,12$ $\text{Given that }(x+3)\text{ and }(x-4)\text{ are factors of }p(x)\text{, factor }p(x)\text{ completely.}$ I'm not sure whether I should sub in -3 and 4 or whether I should divide the original polynomial by x+3 and x-4. [color=blue]Both methods should work . . .[/color]

This is how I solved it . . .

$\text{Since }(x+3)\text{ is a factor, then: }\:p(-3)\:=\:0$
[color=beige]. . [/color]$\text{W\!e have: }\:-27\,+\,9a\,-\,3b\,-\,12 \:=\:0 \;\;\;\Rightarrow\;\;\;\;9a\,-\,3b\:=\:39 \;\;\;\;\;\Rightarrow\;\;\;\; 3a\,-\,b\:=\:13\;\;[1]$

$\text{Since }(x-4)\text{ is a factor, then: }\:p(4) \:=\:0$
[color=beige]. . [/color]$\text{W\!e have: }\:64\,+\,16a\,+\,4b\,-\,12\:=\:0\;\;\;\Rightarrow\;\;\;16a\,+\,4b\:=\:-52 \;\;\;\Rightarrow\;\;\; 4a\,+\,b\:=\:-13\;\;[2]$

$\text{Add [1] and [2]: }\:7a \:=\:0 \;\;\;\Rightarrow\;\;\;a \:=\:0$

$\text{Substitute into [1]: }\:0 - b \:=\:13 \;\;\;\Rightarrow\;\;\;b \:=\:-13$

$\text{Hence: }\:p(x) \:=\:x^3\,-\,13x\,-\,12$

[color=beige]. . [/color]$\text{which factors: }\:p(x) \:=\:(x\,+\,4)(x\,-\,3)(x\,+\,1)$

 April 17th, 2011, 06:06 PM #3 Global Moderator   Joined: Dec 2006 Posts: 21,034 Thanks: 2271 If x³ + ax² + bx - 12 ? (x + 3)(x - 4)(x + r), putting x = 0 gives r = 1.

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