1. ## factorize

Hello guys just a few questions

1.) Factorise completely 2x^3 - 20x^2 + 46x - 28

here i took out the common factor to give 2(x^3 - 10x^2 + 23x - 14)

now when i began trying factors i wasnt sure whether to leave the 2 outside like so: 2((1)^3 - 10(1)^2 + 23(1) - 14))

or just (1)^3 - 10(1)^2 + 23(1) -14

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now the second part is long division... do i leave the two outside when long dividing or do i have to expand i back into the sum?

2. Hi
Originally Posted by andrew2322
now the second part is long division... do i leave the two outside when long dividing or do i have to expand i back into the sum?
You can leave the 2 outside. If you don't want this common factor to appear during the division, you can even factor $x^3-10x^2+23x-14$ and, once it's done, say that $2x^3-20x^2+46x-28=2(x^3-10x^2+23x-14)=\ldots$

Good luck

3. ## hey

me confused =( and another thing, just for general knowledge, if u had something like this : 3(2x-1)^3 -4 could u times the 3(2x-1)^3 by the three first, or do u have to cube it first?

4. Originally Posted by andrew2322
me confused =(
$P(x)=2x^3-20x^2+46x-28=2(x^3-10x^2+23x-14)$ and 1 is a root of this polynomial.

The division of $x^3-10x^2+23x-14$ by $x-1$ will give something like $x^3-10x^2+23x-14=(x-1)(ax^2+bx+x)$ hence $P(x)=2(x^3-10x^2+23x-14)=2(x-1)(ax^2+bx+x)$ : $P$ has been factored too. That's why you only need to factor $x^3-10x^2+23x-14$.

Is it clearer ?

just for general knowledge, if u had something like this : 3(2x-1)^3 -4 could u times the 3(2x-1)^3 by the three first, or do u have to cube it first?
I don't understand what you mean.

5. ## hey

with the 3(2x-1)^3 bit, can u times the 3 in first or do u have to cube (2x-1) first