How do I factor: 2n^2-21-n?

Question

I completely forgot how to factor and it seems like looking it up online doesn't help. Can someone guide me through this step by step?

Answer 1

2n^2 - n - 21

Must be (2n.....)(n(.....)

To get the 21 try 3 and 7: (2n - 7)(n + 3)

outer = 6n
inner = -7n
they add up to -1n or -n

So that's it.

Answer 2

2n^2 - n - 21 = 0

This will always be the first step:
2*(-21) = -42 2 is from 2n^2 and -21 is from -21
x1 + x2 = -1 -1 is from -n which is -1n

notice that 7*6 = 42
since we need -42, it will be either -7*6 or -6*7
but we know the sum of x1 n x2 should be -1
so we choose -7 as x1 and 6 as x2
that will make the multiplication = -42 and the sum = -1

now:
2n^2 - n - 21 = 0
2n^2 + 6n - 7n -21 = 0 ======change -n by the x1 and
x2 that we get earlier
above
(2n^2 + 6n) - (7n+21) = 0 ======group them
2n(n+3) -7(n+3) = 0 ======factor them out
(2n - 7) (n+3) = 0

2n-7 = 0
n = 7/2

n+3 = 0
n= -3

so n1= 7/2 and n2 = -3

hope it will be helpful

Answer 3

The quadratic formula always works. Where the variables are in standard form, 0=ax^2+bx+c [0=2n^2+(-n)+(-21)].
the formula is 0=[-b +/-S QRT(b^2-4ac)]/2a.
0 = [-(-1) +/- SQRT((-1^2)-(4*2*(-21))]/2*2
= [1 +/- SQRT(1 +168)]/4 = [1 +/- SQRT 169]/4
= [1+/-13]/4
= 14/4 and -12/4 = 3.5 and -3
0 = 3.5 so 2n = 7 and n= -3
Therefore 0 = (2n-7)(n+3)

The nice thing about using the quadratic formula is that it works even it the roots are irrational or imaginary, whereas the other answers don't work except on rational roots.

Answer 4

First rewrite 2n^2-21-n in descending order to get 2n^2-n-21. This will then factor as (2n-7)(n+3)

Answer 5

Ok, let's see,

by trial and error you will find

(2n -7)(n+3) = 2n^2 + 6n - 7n - 21
= 2n^2 - n - 21

and that's how is done.

Answer 6

2n^2 - n - 21
(2n - 7) (n + 3)

Answer 7

2n^2-21-n = (2n-7)(n+3)
--------
Ideas: 2(-21) = 6(-7) and 6-7 = -1.

Answer 8

2n^2-21-n
=2n^2-n-21
=2n^2+6n-7n-21
=2n(n+3)-7(n+3)
=(n+3)(2n-7)