Question

How do you factor $20{{x}^{2}}+37x+15$ ?

Answer 1

Hint: This problem on factorization of quadratic expression can be done by using the factors of $20\times 15$ which is $300$ . In this case we will be using the factors $25$ and $12$, then expressing $37$ in the middle term as the sum of $25$ and $12$ like $20{{x}^{2}}+\left( 25+12 \right)x+15$ . After, further simplification we can take $5x$ common from $20{{x}^{2}}+25x$ giving $5x\left( 4x+5 \right)$ and $3$ common from $12x+15$ giving $3\left( 4x+5 \right)$ and again taking the term $\left( 4x+5 \right)$ common we will get the two factors of this expression.

Complete step by step answer:
The given expression we have is
$20{{x}^{2}}+37x+15$
We start by comparing this expression with a general quadratic expression as
$a{{x}^{2}}+bx+c$
To find the factors first we need to find a pair of integers whose product is $ac$
 and whose sum is $b$.
In this case, the product is $20\times 15=300$
Now, the number $300$ can be explained as $30\times 10$, $20\times 15$ , $12\times 25$ and so on.
So, to get $37$ as the coefficient of $x$ in the middle term we will use the factors $25$ and $12$ .
The expression now can be written as
$\Rightarrow 20{{x}^{2}}+\left( 25+12 \right)x+15$
We now apply the distribution on the middle term of the expression and get
$\Rightarrow 20{{x}^{2}}+25x+12x+15$
We can now take $5x$ common from $20{{x}^{2}}+25x$ and the expression becomes
\[\Rightarrow 5x\left( 4x+5 \right)+12x+15\]
Also, we take $3$ common from the rest terms \[12x+15\] and the expression becomes
\[\Rightarrow 5x\left( 4x+5 \right)+3\left( 4x+5 \right)\]
Again, we can see that the term $\left( 4x+5 \right)$ can be taken as common from the entire expression as
$\Rightarrow \left( 4x+5 \right)\left( 5x+3 \right)$
Therefore, we conclude that the factors of the expression $20{{x}^{2}}+37x+15$ are $\left( 4x+5 \right)$ and $\left( 5x+3 \right)$ , and the expression can be factored as $\left( 4x+5 \right)\left( 5x+3 \right)$ .

Note: The problems on factorization of quadratic expression can be done by equating the entire expression with zero and applying the Sridhar Acharya formula to get the solution of the quadratic equation, which is $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . Applying we get $x=-\dfrac{5}{4},\dfrac{3}{5}$ and which leads us to the final factors.