how to factorize $2n^2-n-1$?

Question

How to factorize $2n^2-n-1$?

I solved the equation of the second degree and I get $n_1= 1$ and $n_2= -0.5$.

The solution should be: $(2n+1)(n-1)$

Answer 1

Well, you can check your own answer and see that you are correct. To derive this another way, you can try the following re-write trick that often works:

$$ 2n^2-n-1 = 2n^2 - 2n + n - 1= 2n(n-1) + 1(n-1) = (2n+1)(n-1). $$

Answer 2

Write it as $$2n^2-2n+n-1$$ $$2n(n-1)+1(n-1)$$ $$(2n+1)(n-1)$$ If $$2n+1=0$$ Then $$n=-0.5$$ If $$n-1=0$$, then $$n=1$$

Answer 3

$$2(n^2-0.5n-0.5)=2(n-a)(n-b)$$

$$2(n^2-0.5n-0.5)=2(n^2-(a+b)n+ab)$$

$$a+b= 0.5$$

$$ab= -0.5$$

Squaring the first one $$a^2+b^2+2ab= 0.25$$

Subtract four times the second one

$$a^2+b^2-2ab= 2.25$$

$$(a-b)^2= 2.25$$

$$a-b= \pm1.5$$

Let's continue with plus, as $a$ and $b$ can switch places.

$$a-b= 1.5$$

$$a+b= 0.5$$

Solving

$$a=1, b=-1/2$$

So

$$2(n^2-0.5n-0.5)=2(n-1)(n-(\frac{-1}{2})$$

$$2(n^2-0.5n-0.5)=2(n-1)(n+0.5)$$

$$2(n^2-0.5n-0.5)=(n-1)(2n+1)$$

Answer 4

You have already found the roots $n_1$, and $n_2$ so by definition $$ (n-n_1)(n-n_2)=0 $$ i.e. $$ (n-1)(n+0.5)=0 $$ Clear fractions $$ 2(n-1)(n+0.5)=0=(n-1)(2n+1) $$

An alternative is to guess simple numbers for roots, like $\pm1$.

Trying $n=1$ shows you this is a root. Then you can obtain the other factor by division or writing the following, where $a$ is to be determined and the $2$ is known a-priori because you want a $2n^2$ term. $$ (n-1)(2n+a)=2n^2-2n+an-a $$ and for this to match your quadratic it is necessary that $a=1$

You might want to check the Factor theorem