Factoring equation with 4 terms: 4x³ - 8x² - x + 2.

[Banks]

Factoring equation with 4 terms: 4x³ - 8x² - x + 2.

Hey all,

I'm pretty rusty with my factoring and am trying to factorise: 4x³ - 8x² - x + 2.

I know to split down the middle first then factor and from this I got: 4x² (x - 2) -1 (x - 2) which then became (4x² - 1)(x - 2)²

At this point I don't know what to do. I thought that i'd got it but it didn't match the answer in my book so could someone please advise?

Thanks

 

[Harry_the_cat]

Hey all,

I'm pretty rusty with my factoring and am trying to factorise: 4x³ - 8x² - x + 2.

I know to split down the middle first then factor and from this I got: 4x² (x - 2) -1 (x - 2) which then became (4x² - 1)(x - 2)²

Actually, this becomes (4x² - 1)(x - 2) .... ie. note the (x-2) is not squared.

Compare it to 5a - 2a = (5-2)a =3a ... NOT ...3a².

At this point I don't know what to do. I thought that i'd got it but it didn't match the answer in my book so could someone please advise?

You now need to factorise the first bracket further. (Hint: difference of 2 squares.)

Thanks

See comments in red.

 

[Banks]

See comments in red.

Got it now, thank you! So the -1 (x+2) bit becomes redundant because we subtract one of the (x + 2) factors, so we're left with only one (x + 2) ?

Then after factorising the difference of two squares, we're left with (2x - 1)(2x + 1)(x+2).

 

[MarkFL]

Hey all,

I'm pretty rusty with my factoring and am trying to factorise: 4x³ - 8x² - x + 2.

I know to split down the middle first then factor and from this I got: 4x² (x - 2) -1 (x - 2) which then became (4x² - 1)(x - 2)²

At this point I don't know what to do. I thought that i'd got it but it didn't match the answer in my book so could someone please advise?

Thanks

While this polynomial is easy to factor by grouping, the general method I use for polynomials whose order is greater than 2 is to first try to find one of the rational roots (using the rational roots theorem for the candidates). Let:

\(\displaystyle f(x)=4x^3-8x^2-x+2\)

We can easily see that:

\(\displaystyle f(2)=0\)

So, let's use synthetic division:

\(\displaystyle \begin{array}{c|rr}& 4 & -8 & -1 & 2 \\ 2 & & 8 & 0 & -2 \\ \hline & 4 & 0 & -1 & 0 \end{array}\)

From this we conclude:

\(\displaystyle f(x)=(x-2)\left(4x^2-1\right)\)

And now we can finish factoring by using the difference of squares:

\(\displaystyle f(x)=(x-2)(2x+1)(2x-1)\)