Core 1 Maths problem. Can someone help me understand?

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?

Please help clear it up for me

Reply 1

alex2100x

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?

Please help clear it up for me

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)

(edited 9 years ago)

Reply 2

Muttley79

20

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.

Reply 3

Hasufel

16

"..therefore, by the factor theorem, (x-2) is a root of the given polynomial..."

Reply 4

wednesday_adams

OP

8

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?

Reply 5

Factorisation

8

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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Reply 6

Factorisation

8

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

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Reply 7

alex2100x

2

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 :smile:

Posted from TSR Mobile

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.

Reply 8

alex2100x

2

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

Posted from TSR Mobile

He is dividing by (x+6) so we should expect (x+6) to be a factor of f(x) not (x-6)