Question

How do you factor ${{y}^{2}}-16y+64$?

Answer 1

Hint: To factor the above expression ${{y}^{2}}-16y+64$, we are going to first of all multiply the coefficient of ${{y}^{2}}$ with the constant term and then write the factors of that multiplication. Now, we have to find that combination of the factors which in addition and subtraction will give the coefficient of y. Then we will find some common terms in the expression which we will take out.

Complete step-by-step answer:
The expression in y which is given in the above problem i.e.:
${{y}^{2}}-16y+64$
As you can see the degree of the above expression is 2 because the highest power of the variable in the above expression is 2. Now, we are going to multiply the coefficient of ${{y}^{2}}\left( i.e.1 \right)$ with the constant term (i.e. 64) we get,
$\Rightarrow 64$
Now, we will write the factors of the above number 64 which is equal to:
$\begin{align}
  & 64=1\times 64 \\
 & 64=2\times 32 \\
 & 64=4\times 16 \\
 & 64=8\times 8 \\
\end{align}$
If you carefully speculate the above factors then you will find that if we add the last two factors (i.e. 8 and 8) we get 16 and the coefficient of y in the above expression (is -16) so substituting $\left( 8+8 \right)$ in place of 16 in the above quadratic expression in y we get,
$\Rightarrow {{y}^{2}}-\left( 8+8 \right)y+64$
Now, multiplying y by 8 and then 8 we get,
$\Rightarrow {{y}^{2}}-8y-8y+64$
Taking y as common from the first two terms of the above expression and -8 as common from the last two terms we get,
$\Rightarrow y\left( y-8 \right)-8\left( y-8 \right)$
As you can see that $\left( y-8 \right)$ is common in the above expression so taking $\left( y-8 \right)$ out from the above expression we get,
$\left( y-8 \right)\left( y-8 \right)$
Hence, we have factored the expression given in the above problem as $\left( y-8 \right)\left( y-8 \right)$.

Note: The alternate way of approaching the above problem is as follows:
If you carefully look the expression ${{y}^{2}}-16y+64$ given in the above problem this is the perfect square and we can write this expression as follows:
${{y}^{2}}-2.y.8+{{\left( 8 \right)}^{2}}$
The above expression is the expansion of the following algebraic identity:
${{\left( a-b \right)}^{2}}={{a}^{2}}-2.a.b+{{b}^{2}}$
Now, substituting $a=y$ and $b=8$ in the above equation we get,
$\Rightarrow {{\left( y-8 \right)}^{2}}={{y}^{2}}-2.y.8+{{8}^{2}}$
Using the above relation, in ${{y}^{2}}-2.y.8+{{\left( 8 \right)}^{2}}$ we get,
$\Rightarrow {{\left( y-8 \right)}^{2}}$
We know that square of any expression is the multiplication of same expression twice so we can write as:
$\left( y-8 \right)\left( y-8 \right)$