Question

How do you factor the trinomial ${p^2} - 2p + 1$?

Answer 1

Hint: This equation is the quadratic equation. The general form of the quadratic equation is $a{x^2} + bx + c = 0$. Where ‘a’ is the coefficient of ${x^2}$, ‘b’ is the coefficient of x and ‘c’ is the constant term.
To solve this equation, we will apply the sum-product pattern. During the simplification, we will take out common factors from the two pairs. Then we will rewrite it in factored form.
Therefore, we should follow the below steps:
> Apply sum-product pattern.
> Make two pairs.
> Common factor from two pairs.
> Rewrite in factored form.

Complete step-by-step answer:
Here, the quadratic equation is
$ \Rightarrow {p^2} - 2p + 1$
Let us apply the sum-product pattern in the above equation.
Since the coefficient of ${p^2}$is 1 and the constant term is 1. Let us multiply 1 and 1. The answer will be 1. We have to find the factors of 1 which sum to -2. Here, the factors are -1 and -1.
Therefore,
$ \Rightarrow {p^2} - p - p + 1$
Now, make two pairs in the above equation.
$ \Rightarrow \left( {{p^2} - p} \right) - \left( {p - 1} \right)$
Let us take out the common factor.
$ \Rightarrow p\left( {p - 1} \right) - 1\left( {p - 1} \right)$
Now, rewrite the above equation in factored form.
$ \Rightarrow \left( {p - 1} \right)\left( {p - 1} \right)$
That is equal to
$ \Rightarrow {\left( {p - 1} \right)^2}$

Note:
One important thing is, we can always check our work by multiplying out factors back together, and check that we have got back the original answer.
To check our factorization, multiplication goes like this:
$ \Rightarrow \left( {p - 1} \right)\left( {p - 1} \right)$
Let us apply multiplication to remove brackets.
$ \Rightarrow {p^2} - p - p + 1$
Let us simplify it. We will get,
$ \Rightarrow {p^2} - 2p + 1$
Hence, we get our quadratic equation back by applying multiplication.
Here is a list of methods to solve quadratic equations:
> Factorization
> Completing the square
> Using graph
> Quadratic formula