# Quick question on factor theoremWatch

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#1
use the factor theorem to factorise the following cubics as far as possible:
f(x) = x^3 -7x-6

As there is no x^2 in the cubic do i need to first do algebraic division?
0
6 years ago
#2
(Original post by dippers)
use the factor theorem to factorise the following cubics as far as possible:
f(x) = x^3 -7x-6

As there is no x^2 in the cubic do i need to first do algebraic division?
See what the values of f(1) f(-1) etc You are looking for the value to be zero. This gives you your first factor.
0
6 years ago
#3
If f(a) =0, then the algebraic expression has the factor (x-a). That's the factor theorem. Experiment with values of 1,-1,2,-2 etc to see what the factor is. Once you find this factor you can use algebraic division to find the others.
0
#4
(Original post by steve2005)
See what the values of f(1) f(-1) etc You are looking for the value to be zero. This gives you your first factor.
I did that then I did algrebraic division and got x^2-x-6, is this right? Then I'm not sure how to factorise it. I know you use (x+1) for the first bracket and i'm unsure how to do the rest.
0
6 years ago
#5
(Original post by dippers)
I did that then I did algrebraic division and got x^2-x-6, is this right? Then I'm not sure how to factorise it. I know you use (x+1) for the first bracket and i'm unsure how to do the rest.

You could do long division OR synthetic division.

0
#6
I'm having problems with this question
Find a and b given that (x-1) and (2x+3) are factors of f(x)= ax^3+3x^2+bx-3.
I know I input 1 and -3/2 into the equation and then solve by doing simulataneous equations but I keep getting the wrong answer.
0
6 years ago
#7
(Original post by dippers)
I'm having problems with this question
Find a and b given that (x-1) and (2x+3) are factors of f(x)= ax^3+3x^2+bx-3.
I know I input 1 and -3/2 into the equation and then solve by doing simulataneous equations but I keep getting the wrong answer.
do you want to show your working
0
#8
(Original post by TenOfThem)
do you want to show your working
x=1 0=a+b and x=-3/2 -15/4=27/8a+3/2b
0
6 years ago
#9
(Original post by dippers)
x=1 0=a+b and x=-3/2 -15/4=27/8a+3/2b
they are correct
0
#10
(Original post by TenOfThem)
they are correct
How do i solve it from then, do i change the 0=a+b equation by multiplying by 3/2?
0
6 years ago
#11
(Original post by dippers)
How do i solve it from then, do i change the 0=a+b equation by multiplying by 3/2?
There are many approaches to solving simultaneous equations

Any method should be fine

Including the one you suggest
0
#12
(Original post by TenOfThem)
There are many approaches to solving simultaneous equations

Any method should be fine

Including the one you suggest
I've done it now, thank you!
0
6 years ago
#13
(Original post by dippers)
I still keep getting the wrong answer, i'm not sure why.
What do you think the correct answer is

What do you get
0
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