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Functions (factorising) question

I'm having some trouble with this question to do with factorising a cubic equation... I have no idea what to do with the given information.

Thanks in advance

Reply 1

NotNotBatman

The factor theorem is; if f(a)=0, then (x-a) is a factor of f(x), this would mean (x-a)(some quadratic factor) = 0. You can use algebraic long division to solve for the quadratic factor.

Reply 2

BobBobson

Original post by Gabzinc

I'm having some trouble with this question to do with factorising a cubic equation... I have no idea what to do with the given information.

Thanks in advance

If f(-4) =0 you know that x+4 is a factor of it. Start by factorising that out

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Reply 3

Gabzinc

Original post by NotNotBatman

The factor theorem is; if f(a)=0, then (x-a) is a factor of f(x), this would mean (x-a)(some quadratic factor) = 0. You can use algebraic long division to solve for the quadratic factor.

Original post by BobBobson

If f(-4) =0 you know that x+4 is a factor of it. Start by factorising that out

I've never studied factor theorem or factorising cubics before. I'll give it some practice tonight.

Thanks guys.

Reply 4

username1842595

Original post by Gabzinc

I'm having some trouble with this question to do with factorising a cubic equation... I have no idea what to do with the given information.

Thanks in advance

Since you've been given that x=-4 is a root that must mean you can write the polynomial in the form:

(x + 4) times a quadratic. Simple log division will give you the quadratic:

f(x) = (x+4)(2x^2 -x -3)

Now you just have to factor the quadratic. Try a few things. You should quickly discover that the full factorization is:

f(x) = (x+4)(x+1)(2x-3)

So the roots of f(x) = 0 are -4, -1, 3

Hopefully this helps. :smile:

Reply 5

Mr M

Study Forum Helper

Original post by sameehaiqbal

Hopefully this helps. :smile:

Unfortunately providing a full solution rarely helps and that is why it is frowned upon here. Please read the Forum Rules pinned to the top of the Maths Forum.

Reply 6

username1842595

Original post by Mr M

Unfortunately providing a full solution rarely helps and that is why it is frowned upon here. Please read the Forum Rules pinned to the top of the Maths Forum.