Question

How do you factor `x²-11x+30`?

Answer 1

`(x-5)(x-6)`
Refer to explanation.

Explanation:

This is a quadratic equation in the form of `ax^2+bx+c`, so you simply factor b into two parts such that their sum is b and product is ac.

`x^2-11x+30`
`x^2-5x-6x+30`
`( -5xx-6=30 and -5-6=-11)`
`x(x-5)-6(x-5)`
`(x-5)(x-6)`

That's it.
Hope this helps :)

Answer 2

`x^2-11x+30=color(blue)((x-5)(x-6))`

Explanation:

If we attempt to factor `x^2-11x+30`
into the product of two binomials: `(x+a)(x+b)`
since `(x+a)(x+b)=x^2+(a+b)x+ab`
we see that
`a` abd `b` must have the same sign (since the constant term is `color(red)+30`)
`ab` must be equal to `30`
`a+b` must add up to `-11` (which implies they both must be negative).

#{: ("negative factor pairs of "30":"," | ",-1xx-30," | ",-3xx-10," | ",color(blue)(-5xx-6)), ("sum of factor pairs:"" | ",,-31," | ",-13," | ",color(blue)(-11)) :}#

Answer 3

`x=5` `,6`

Explanation:

On comparing this equation with the equation `ax^2+bx+c = 0`
`a = 1,b= -11,c= 30`
Now you have to split the middle term i.e. -11 in two parts such that the multiplication of both the parts is equal to `a*c` while the addition of both the parts is equal to `b` itself.
We can divide `-11` as `-6` and `-5`
So, the equation becomes
`(x^2 +(-6-5)x+30) = 0`
`(x^2-6x-5x+30) = 0`
Now taking out common terms
`(x[x-6]-5[x-6]) = 0`
`([x-5][x-6]) = 0`
`x=5 x=6`