Question

How do you factor $9{{x}^{2}}+30x+25$?

Answer 1

Hint: To find the factors of the given polynomial we will use the algebraic identities. We know that ${{\left( ax+b \right)}^{2}}={{a}^{2}}{{x}^{2}}+{{b}^{2}}+2abx=\left( ax+b \right)\left( ax+b \right)$, so we compare the coefficients of the given polynomial with the identity and find the factors.

Complete step by step answer:
We have been given a polynomial $9{{x}^{2}}+30x+25$.
We have to find the factors of the given polynomial.
Now, we can write the terms of the given polynomial as
$\begin{align}
  & 9{{x}^{2}}={{\left( 3x \right)}^{2}}, \\
 & 25={{5}^{2}}, \\
 & 30x=2\times 3\times 5x \\
\end{align}$
Now, we know that ${{\left( ax+b \right)}^{2}}={{a}^{2}}{{x}^{2}}+{{b}^{2}}+2abx=\left( ax+b \right)\left( ax+b \right)$
When we compare the given polynomial with the above explained identity, we will get
$\begin{align}
  & a=3, \\
 & b=5 \\
\end{align}$
So we can write the given polynomial in factors form as $\left( ax+b \right)\left( ax+b \right)$ by using the algebraic identity.
Now, substituting the values we will get
$\Rightarrow \left( 3x+5 \right)\left( 3x+5 \right)$

Hence we get the two factors of the given polynomial $9{{x}^{2}}+30x+25$ as $\left( 3x+5 \right)\left( 3x+5 \right)$.

Note: We have different methods to factorize a polynomial. Alternatively, we can also factorize the given polynomial by splitting the middle term. We can also use grouping method, sum or difference in two cubes, greatest common factor (GCF) method, trinomial method etc.
Here in this question we can split the middle term by using sum-product form and then write common factors from the two pairs.
We have $9{{x}^{2}}+30x+25$
Now, using sum-product form we can write the above polynomial as
$\Rightarrow 9{{x}^{2}}+15x+15x+25$
Now, taking common terms out we will get
$\Rightarrow 3x\left( 3x+5 \right)+5\left( 3x+5 \right)$
Now, we can write the above obtained equation as
$\Rightarrow \left( 3x+5 \right)\left( 3x+5 \right)$
So, we get two common factors of the given polynomial.