Question

Explain how to factor 2x^2 + 7x - 4 = 0

Answer 1

(mx+p)(nx+q)=mnx^2+mqx+pnx+pq = (mn)x^2 +(mq+pn)x + (pq) = ax^2+bx+c.
Working backwards from that, find factors of a*c = m*n*p*q, such that b=mq+pn.
In other words, find factors of (2)(-4) that add up to 7.

Answer 2

how do you get from 2x^2 + 8x - x -4 = 0 tp 2x(x+4)-1(x+4) = 0 ?

Answer 3

to*

Answer 4

That's called 'factor-by-grouping' - factor the first two terms then the last two terms, then find a common factor in those.

Answer 5

how come it isn't (x+4)^2 since there's two of them

Answer 6

It would only be (x+4)^2 if they were being multiplied, but they are being added here.

Answer 7

I don't understand where are they being added?

Answer 8

Assume that u have an equation \(ax^2+bx+c\). then,
u must break the middle term (bx) such that the product of middle terms become the product of (c & ax^2) & after add in them we get (bx)

Answer 9

@monk3yboy305 :)

Answer 10

Or 'subtracted' if you like
"2x(x+4)-1(x+4) "
The two terms are 2x(x+4) and -1(x+4) and they are being added together. (Or were, now we are splitting them apart).

Answer 11

It's like if you had 6 + 24. The common factor is 6, so you can rewrite it as 6(1+4).

Answer 12

Oh okay I undestand :) Thanks both of you

Answer 13

Assume that u have an equation like \(\Large{ax^2+bx+c}\).then, to factor the equation we must use the identity \[\Huge{\color{red}{x^2+(a+b)x+c}}\]