x^3-3x^2-x+3 ... factor

what do you get when you factorise x^3-x?

@Laurenn_marie7

x(x^2-1) ? @ajprincess

ya good:) Nw can u factorise -3x^2+3?

3(x^2-1)

well small correction. it is -3(x^2-1)

Nw can u factorise x(x^2-1)-3(x^2-1)?

that's the part i need help on!

think x^2-1 as y. Then u will have xy-3y Can u factorise this?

so factor out x^2-1 or just -1?

factor out x^2-1

x^2-1(x-3)

ya gud:) Nw x^2-1 can still be factoried. Can u do it? It is difference of squares.

yes got it, thank you!! x-1 and x+1. thanks for all of the help! can you help me out with (2x-5)^4 ?

@ajprincess

ya sure:) Well what do u want to do with it?

factor it out! so it would be (2x-5)(2x-5)(2x-5)(2x-5) right?

ya

and then I got (4x^2-10x-10x+25)(4x^2-10x-10x+25) and I'm stuck there

do u want to distribute it?

Yes to just factor completely

factor completely will yield jst (2x-5)(2x-5)(2x-5)(2x-5)

if u want to distribute use pascal's triangle

what is that?

here i dont know how to explain it ill give you the answer 192/125

it a fraction

no the answer is 16x^4-160x^3+600x^2+625 but I'm not sure how they got all of that

i thought that to but i went with that one

well how do you get that from (4x^2-20x+25)(4x^2-20x+25) ?

you right i factored it wrong sorry

pascal, triangle |dw:1363107248021:dw| The numbers in the pascal's triangle are the coefficients of the terms that make up the expression that u want to distribute. If u have an expression with an exponent then there will be n+1 terms when u distribute it. In ur case u have (2x-5)^4 since the exponent is 4 there will be 5 terms. nw c1(2x)^4+c2(2x)^3*5+c3(2x)^2*5^2+c4(2x)*5^3+c55^4 C1,c2,c3,c4 and c5 are the coefficients. frm the pascal's triangle we can find that they c1=1, c2=4,c3=6,c4=4 and c5=1. so c1(2x)^4+c2(2x)^3*5+c3(2x)^2*5^2+c4(2x)*5^3+c55^4 becomes (2x)^4-4*(2x)^3*5+6*(2x)^2*5^2-4*(2x)*5^3+5^4 =2^4x^4-4*2^3x^3*5+6*2^2x^2*5^2-4*2x*5^3+5^4 =16x^4-160x^3+600x^2-1000x+625

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