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# Core 1 Maths problem. Can someone help me understand? watch

1. Hi, this question is on polynomials revision booklet.
It says:

You are given that f(x) = x^3 + 9x^2 + 20x + 12

i) Show that x=-2 is a root of f(x) = 0

My question is, how do I show that something is a root?
Do I just sub f(2) into the equation and get 2^3 + 9(2^2) + 20(2) + 12 = 0
?
but then what would be the next step to show its a root?

I'm getting confused between the terms "factor" and "root" what is the difference?

2. (Original post by wednesday_adams)
Hi, this question is on polynomials revision booklet.
It says:

You are given that f(x) = x^3 + 9x^2 + 20x + 12

i) Show that x=-2 is a root of f(x) = 0

My question is, how do I show that something is a root?
Do I just sub f(2) into the equation and get 2^3 + 9(2^2) + 20(2) + 12 = 0
?
but then what would be the next step to show its a root?

I'm getting confused between the terms "factor" and "root" what is the difference?

Yes you just plug in x=-2 and show that it is zero as you have done. A root is x=-2 in this case. A factor is (x+2) as if a is a root of f(x)=0 then (x-a) is a factor of f(x).

Consider that you can write f(x) as (x-a)(x-b)(x-c) if f(x) has 3 roots so each of (x-a),(x-b) and (x-c) is a factor of f(x) since some multiple of it is equal to f(x)
3. Once you have made the substitution and got f(-2) = 0 then you mst state that therefore x = -2 is a root (or solution). Just substituting on its own is not enough.
4. "..therefore, by the factor theorem, (x-2) is a root of the given polynomial..."
5. (Original post by alex2100x)
Yes you just plug in x=-2 and show that it is zero as you have done. A root is x=-2 in this case. A factor is (x+2) as if a is a root of f(x)=0 then (x-a) is a factor of f(x).

Consider that you can write f(x) as (x-a)(x-b)(x-c) if f(x) has 3 roots so each of (x-a),(x-b) and (x-c) is a factor of f(x) since some multiple of it is equal to f(x)
thank you! I understand where I went wrong now. I was plugging in f(2) rather than -2.

But part 2 says:

(ii) Divide f(x) by x+6

and I have no idea how to do this..

do I plug f(-6) in or something? but how?
6. You should be able to factorise f(x) into brackets. One of those brackets should be (x-6) and therefor you can cancel the two brackets out

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7. I should add... Add c1 if youre asked to factorise a cubic, just take an x out and factorise the quadratic

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8. (Original post by Factorisation)
I should add... Add c1 if youre asked to factorise a cubic, just take an x out and factorise the quadratic

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Taking the x out will not help him as there is a term independent of x, namely the +12. See my other post on polynomial long division on the other thread OP.
9. (Original post by Factorisation)
You should be able to factorise f(x) into brackets. One of those brackets should be (x-6) and therefor you can cancel the two brackets out

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He is dividing by (x+6) so we should expect (x+6) to be a factor of f(x) not (x-6)
10. (Original post by alex2100x)
He is dividing by (x+6) so we should expect (x+6) to be a factor of f(x) not (x-6)
Oops sorry, my bad!

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11. (Original post by Hasufel)
"..therefore, by the factor theorem, (x-2) is a root of the given polynomial..."
be careful....

the root is -2

the factor is ( x + 2 )
12. (Original post by the bear)
be careful....

the root is -2

the factor is ( x + 2 )
s**t! - my bad! -i hate myself for that! - sorry fellow posters!

13. (Original post by Hasufel)
s**t! - my bad! -i hate myself for that! - sorry fellow posters!

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Updated: December 29, 2014
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