Question

How do you factor completely $9{a^2} - 24ab + 16{b^2}$?

Answer 1

Hint: In this question we are asked to factor the equation. Note that the given equation is a quadratic equation. If we carefully observe the equation some terms are perfect squares. So rewrite them as squares of some number or variable. To solve this we make use of the perfect square trinomial rule which is given as ${(x - y)^2} = {x^2} - 2xy + {y^2}$. Substitute the values of the variables x and y and the solve to obtain the required result.

Complete step by step answer:
Given an equation of the form,
$9{a^2} - 24ab + 16{b^2}$ …… (1)
Here we are asked to factor the above equation (1).
Note that some of the terms in the equation are the perfect squares of some number.
So we rewrite them and solve the problem.
Rewrite $9{a^2}$ as ${(3a)^2}$.
Rewrite $16{b^2}$ as ${(4b)^2}$.
So the equation (1) becomes,
$ \Rightarrow {(3a)^2} - 24ab + {(4b)^2}$ …… (2)
Note that the above equation is of the form of a perfect square trinomial.
So we make use of the perfect square trinomial rule to solve the equation.
The perfect square trinomial rule is given by,
${(x - y)^2} = {x^2} - 2xy + {y^2}$
Note that here $x = 3a$ and $y = 4b$.
Now we check the middle term by multiplying x and y by 2 and this results with the middle term in the original equation.
$2xy = 2 \cdot 3a \cdot 4b$
$ \Rightarrow 2xy = 24ab$
Hence the perfect square trinomial rule is satisfied.
Therefore, the equation (2) becomes,
$ \Rightarrow {(3a - 4b)^2}$.

Hence, the factor of the equation $9{a^2} - 24ab + 16{b^2}$ is ${(3a - 4b)^2}$.

Note: Students must know the perfect squares of the numbers. Otherwise it becomes difficult to solve this kind of problem.
Also they must know the perfect square trinomial rule to find the factors of a given equation.
We recognise the perfect square trinomial rule by the following facts.
(1) It contains three terms.
(2) Two of its terms are perfect squares themselves.
(3) The remaining term will be twice the product of the square root of the other two terms.