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the factor theorem maths help

f(x)=x^3-x^2-14x+24
a) show that x+4 is a factor of f(x)
b) factorise f(x)
c) solve the equation f(x)=0.


pls help im stuck
Original post by M12S
f(x)=x^3-x^2-14x+24
a) show that x+4 is a factor of f(x)
b) factorise f(x)
c) solve the equation f(x)=0.


pls help im stuck


a) If x-a is a factor, then f(a) = 0. Using this, you can show x+4 is a factor.
b) Use polynomial division to divide by x+4. Then factorise what's left. Alternatively, use the same approach as in part (a) with trial and error for random numbers (try -1, 1, 2, -2...)
c) Follows from your answer to b. If (x-a)(x-b)(x-c) = 0 then x=a, b, c.
(edited 8 years ago)
a) Sub x into the equation, e.g f(-4)=-4^3 and so on. If it is a factor, f(x)=0
Hi there


(a) The Factor Theorem is such that if you have some factor, (ax+b), then f(-b/a) = 0. Use this to show (x+4) is a factor.
(You may also come across the remainder theorem, where f(-b/a) is the remainder when f(x) is divided through by (ax+b)



(b) So now that you have shown that (x+4) is a factor. Factorising a cubic may look tricky at first ... but you'll soon see that it is fairly straightforward.
To factorise a cubic, it would be in the form: (ax+b)(cx^2 + dx + e) ... since you would get an x^3 term if you expand this.

Using part (a), you should have the (ax+b) bit ... which is (x+4). If you divide f(x) by this factor by doing some algebraic long division, you'll obtain the quadratic (cx^2 + dx + e). Then see if this can be factorised further as you normally would with a quadratic


(c) Having factorised f(x), let each bracket equal 0 and solve

If you get stuck again, let me know! :smile:
i've kind of answered the questions for you and tried putting in annotations to understand what i've done- i hope this helps


a) If x-a is a factor then f(a)=0

f(-4)=(-4)^3-(-4)^2-14(-4)+24
= 0
therefore (x- -4) = (x+4) is a factor



b) since (x+4) is a factor, (use polynomial knowledge or algerbraic division)

(x+4) ( x^2 +bx +c) = x^3 + bx^2 + cx +4c
+4x^2 +4bx

b+4= -1
therefore b=-5

c+4b=-14
c+ 4 x -5 =-14
c=6

subst. values

(x+4) (x^2 -5x + 6)

factorise

(x+4) (x-2) (x-3) are factors of f(x)



c) since f(x)=0
(x+4) (x-2) (x-3) = 0

x=-4, x=2, x=3

if you need any help or get stuck lettme know :biggrin:
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dangerbby
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Thank you for all the replies.

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