AS Maths Help Factorising

officialzak_2001

can someone show me the steps to 3x^(3)+7x^(2)-4

Reply 1

ManLike007

Original post by officialzak_2001

can someone show me the steps to 3x^(3)+7x^(2)-4

Through trial and error, I found that

x=-1

is a factor (because the answer is zero),

Then you can do

(x+1)(ax^{2}+bx+c)=3x^{3}+7x^{2}+0x^{2}-4

Equate the terms to find

a, b, c

then factorise

ax^{2}+bx+c

.

Another method is long division.

Reply 2

officialzak_2001

Original post by ManLike007

Through trial and error, I found that

x=-1

is a factor (because the answer is zero),

Then you can do

(x+1)(ax^{2}+bx+c)=3x^{3}+7x^{2}+0x^{2}-4

Equate the terms to find

a, b, c

then factorise

ax^{2}+bx+c

.

Another method is long division.

Thanks i was looking for an easier way as this topic is before dividing polynomials

Reply 3

ManLike007

Original post by officialzak_2001

Thanks i was looking for an easier way as this topic is before dividing polynomials

Oh ok, then what you can do is expand out

(x+1)(ax^{2}+bx+c)

and find

a,b,c

so if I had

ax^2+bx+c \equiv 5x^{2}+7

then I know

a=5

b=0

and

c=7

just by looking at both sides and comparing them.

Original post by ManLike007

And just to add another method since OP already knows how to do it, once you find out that

x+1

is a factor:

3x^3+7x^2-4 \equiv 3x^3+3x^2+4x^2+4x-4x-4 \equiv 3x^2(x+1)+4x(x+1)-4(x+1) \equiv (x+1)(3x^2+4x-4)

(this can still be factorised further)

Reply 6

thekidwhogames

Use long division/identities as you know (x+1) is a factor.

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AS Maths Help Factorising

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