How do you factor #x^2 - 22x + 120#?

2 Answers
Mar 25, 2016

Answer:

(x - 10)(x - 12)

Explanation:

#y = x^2 - 22x + 120.#

Standard form #y=ax^2+bx+c#

Use the new Transforming Method (Google, Yahoo)
Find 2 numbers knowing sum (- 22 = b) and product (ac = 120). The numbers have same sign because ac > 0.
Factor pairs of (120) --> ...(6, 20)(10, 12). This last sum is 22 = -b. Then the opposite sum (-10, -12) gives the 2 numbers: -10 and -12 (sum = b)
y = (x - 10)(x - 12)

May 22, 2017

Answer:

Using reasoning:

#x^2-22x+120" "=" "(x-10)(x-12)#

Explanation:

Lets try and reason this out.

The coefficient of #x^2# is +1 so as a starting point we have the form

#( x+-?_1)(x+-?_2)#

The constant of 120 is positive So both of the ? are the same sign.

The #x# term has negative 22 so both ? are negative giving

#(x-?_1)(x-?_2)#

Note that #sqrt(120)->#just under 11

Stating the obvious, the constant of 120 ends in zero so it has to be the product of two numbers where one of them ends in either 5 or 0.

As the square root is close to 11 lets consider the possibility one of the factors as being 10.

#10xx12=120 and (-10)+(-12) = -22" " larrcolor(red)(" It works")#

So we have:

#x^2-22x+120" "=" "(x-10)(x-12)#