# Day 50 – Solving Quadratics by Factoring

s^2\,-\,2s\,-\,35\,=\,0

We’re asked to solve for s.

The best way to solve this as it’s equal to 0 is to factor the left-hand side, and then think about the fact that those binomials that you factor into, that they have to be equal to 0.

There’s also a shortcut if you have 1 as coefficient – s2.

1 – Things about numbers that would be equal to -2 when added, but -35 when multiplied.

a\,+\,b\,=\,-2\\a\,*\,b\,=\,-35\\5\,+\,-7\,=\,-2\\5\,*\,-7\,=\,-35

2 – Now we group it.

s^2\,+\,5s\,-\,7s\,-35\,=0\\s(s+5)\,-7(s+5)\,=\,0

3 – Now we have two terms with (s+5) as a factor, undistributing s+5.

(s+5)(s-7)\,=\,0

4 – Now, either or both these two terms is/are equal to 0.

s+5=0\\s\cancel{+5-5}=\cancel0-5\\s=-5\\s-7=0\\s\,\cancel{-7+7}\,=\,\cancel0\,+7\\s=7

## The Formula

A*B=0\\(x+a)(x+b)\\x^2+bx+ax+ab\\x^2+(a+b)x+ab

Following the formula above, we can just do straight:

(s+5)(s-7)

This entry was posted in Study Notes and tagged , , , , , , . Bookmark the permalink.