Question

How do you factor `2n ^2 - 7n - 4`?

Answer 1

We can factor this as:

`(2n + 1)(n - 4)`

Answer 2

This doesn't factorize by any rules......
so notice that.....
`2n^2=2xxn^2 " or "2nxxn`
Let's just keep it at bay for now

The last term..... `-4` can be
`diamond 1xx-4`
`diamond 4xx-1`
`diamond 2xx-2`

Now come to the bay... `;)`
Notice that if it was `2xxn^2` then it would be
`(2+-x)(n^2+-y)`
Then..... the `x` would also end up getting squared....
And there's only one term squared in the expression, i.e. `2n^2`
So... possibility eliminated
`2n^2=cancel(2xxn^2) " or " 2nxxn`
`2n^2=2nxxn`
Now we get
`(2n+-x)(n+-y)`

Now... as anything isn't coming in my mind... you'll have to consider all the three `diamond` possibilities
`(2n+1)(n-4)`
`2n^2-8n+n-4`
`2n^2-7n-4`

Which... surprisingly.. gives us our answer in our first try

Hope you liked the way and understood well

Answer 3

`(2n+1)(n-4)`

Explanation:

We need to find factors of `2 and 4` whose products subtract to give `7`

(The fact that the third term is negative indicates the subtraction)

Note that the biggest product of the factors of `2 and 4` is `8`. This is our first clue.
Write down the factors and cross multiply:
There can be some trial and error, try different combinations.

`color(white)(xxxx)2" and "4`
`color(white)(xx.x)darrcolor(white)(xxx.x)darr`
`color(white)(xxxx)2color(white)(xxx.xx)1" "rarr 1 xx 1 =1`
`color(white)(xxxx)1color(white)(xx.xxx)4" "rarr 2xx4 ul(=8)`
`color(white)(xxxxxxxxxxxxxx,x....xxxx)7" "larr` subtract

We have the correct factors, now look at the signs.
We need to have `-7,` there are more negatives.

`color(white)(xxxx)2" and "4`
`color(white)(xx.x)darrcolor(white)(xxx.x)darr`
`color(white)(xxxx)2color(white)(xxxxxx)1" "rarr 1 xx 1 =+1`
`color(white)(xxxx)1color(white)(xxxxxx)4" "rarr 2xx4 ul(=-8)`
`color(white)(xxxxxxxxxxxxxx,xxxxxxxx)-7" "larr` negative

Insert the same signs in the second column of factors:

`color(white)(xxxx)2" and "4`
`color(white)(xx.x)darrcolor(white)(xxx.x)darr`
`color(white)(xxxx)color(blue)(2color(white)(xxxx)+1)" "rarr 1 xx +1 =+1`
`color(white)(xxxx)color(red)(1color(white)(xxxx)-4)" "rarr 2xx-4 ul(=-8)`
`color(white)(xxxxxxxxxxxxxx,x.xxxxxxxx)-7" "larr` correct

The rows now give us the factors we need.

`(color(blue)(2n+1))(color(red)(1n-4))`

Multiplying out will confirm that these are the correct factors:

`2n^2 -8n+n -4`

`=2n^2-7n-4`

Answer 4

`2n^2-7n-4 = (2n+1)(n-4)`

Explanation:

Given:

`2n^2-7n-4`

We can use an AC method:

Find a pair of factors of `AC=2*4 = 8` which differ(*) by `B=7`

The pair `8, 1` works in that `8*1 = 8` and `8-1=7`

Use this pair to split the middle term and factor by grouping:

`2n^2-7n-4 = (2n^2-8n)+(n-4)`

`color(white)(2n^2-7n-4) = 2n(n-4)+1(n-4)`

`color(white)(2n^2-7n-4) = (2n+1)(n-4)`

(*) We look for a pair of number whose difference is `B` as opposed to whose sum is `B` because the sign of the constant term is negative. If the constant term were positive, then we would look for a pair of factors whose sum is `B`.