Question

How do you write ${q^2} - 12q + 36$ in factored form?

Answer 1

Hint: First of all, the given expression is a polynomial of degree $2$ . Also, the given expression contains three terms. Now, to write the given expression in factored form, we will use splitting the middle term method. And this way we can solve it further.

Complete step-by-step solution:
We need to factorize the given expression. We know that the given expression is a polynomial in the variable $q$ . Also, the exponent of the highest degree term in a polynomial is known as its degree. Therefore, we can see that the degree of the given polynomial is $2$ .
The polynomial of degree $2$ is called a quadratic polynomial.
Now, let us take the given expression as
$f(q) = {q^2} - 12q + 36$ .
We represented the given expression as $f(q)$ . Also, the given expression is a quadratic polynomial.
Now, to factorize $f(q)$ , we have to find two numbers $a$ and $b$such that $a + b = - 12$ and $ab = 36$ . So clearly, $- 6 - 6 = - 12$ and $- 6 \times - 6 = 36$ .
So, we write the middle term $- 12q$ as $- 6q - 6q$
∴ ${q^2} - 12q + 36$
$= {q^2} - 6q - 6q + 36$
Now. from the first two terms we took out $q$ as common and from the next two terms we took out $- 6$ as common.
$= q(q - 6) - 6(q - 6)$
$= (q - 6)(q - 6)$ , now taking out as common $(q - 6)$ from both the terms

${q^2} - 12q + 36 = (q - 6)(q - 6)$ , which is the required factored form.

Note: The given expression is a quadratic polynomial because the degree of the given polynomial is $2$ . The name ‘quadratic’ has been derived from ‘quadrate’, which means ‘square’. Generally, any quadratic polynomial in variable $x$ with real coefficients is of the form $f(x) = a{x^2} + bx + c$ , where $a$ , $b$ and $c$ are real numbers and $a \ne 0$ .