Question

How do you factor \[{x^2} - 9x + 20\] ?

Answer 1

Hint: In algebra, a quadratic equation is any equation that can be rearranged in standard form as $ax^{2}+bx+c=0$ where x represents an unknown, and a, b, and c represent known numbers, where a ≠ 0. If a = 0, then the equation is linear, not quadratic, as there is no term.

Complete step-by-step answer:
Given, the equation \[{x^2} - 9x + 20\]
First, we check the coefficients of \[{x^2}\] and \[x\] .
Here, the coefficient of \[{x^2}\] is \[1\] , the coefficient of \[x\] is \[ - 9\] .
We now multiply coefficients of \[{x^2}\] and \[20\] . That is,
  \[1 \times 20 = 20\] .
Now, we check for the factors of \[20\] which we can add up to get \[ - 9\] .
Two such factors of \[20\] which can add up to get \[9\] are \[4\] and \[5\]
To get \[ - 9\] , we take the factors as \[ - 4\] and \[ - 5\] .
Now, we write \[\left( { - 4 - 5} \right)\] in place of \[ - 9\] . That is,
  \[{x^2} + \left( { - 4 - 5} \right)x + 20\]
  \[ = {x^2} - 4x - 5x + 20\]
We take \[x\] common from the first two terms and \[5\] common from the last two terms. That is,
  \[{x^2} - 4x - 5x + 20 = x\left( {x - 4} \right) - 5\left( {x - 4} \right)\]
  \[ = \left( {x - 4} \right)\left( {x - 5} \right)\]
Therefore, after factoring \[{x^2} - 9x + 20\] we get \[\left( {x - 4} \right)\left( {x - 5} \right)\] .
So, the correct answer is “ \[\left( {x - 4} \right)\left( {x - 5} \right)\] ”.

Note: By the factorization we mean to break a number into parts which can be multiplied to get the original number. In such types of questions, students mostly make mistakes in plus and minus sign, so while factoring an equation, check carefully the sign of the coefficient of the middle term and whether the factors of the last term add up to give the coefficient of the middle term. Also, remember if the last term is negative that means one factor will be positive and the other will be negative and if the last term is positive that means either both factors will be positive or both factors will be negative.