Question

How do you factor $4{{y}^{2}}+25y+6$?

Answer 1

Hint: We are given a quadratic equation which has to be solved by the method of factoring the equation. We shall break down the y-variable term into two parts which must add up to 25y given in the equation. Then, we will group the common terms and form two linear equations in y-variable out of the given quadratic equation. After further equating each of the linear equations to zero to get our required solution.

Complete step by step solution:
There are four methods for solving quadratic equations, namely, factoring method, completing the square method, taking the square root method and the last method is solving using the various properties of polynomials.
However, we will use the method of factoring the quadratic equation which makes our calculations simpler.
For any quadratic equation $a{{x}^{2}}+bx+c=0$,
the sum of the roots $=-\dfrac{b}{a}$ and the product of the roots $=\dfrac{c}{a}$.
Thus, for the equation, $4{{y}^{2}}+25y+6$,
$\Rightarrow 4{{y}^{2}}+25y+6=0$
We will find numbers by hit and trial whose product is equal to $4\times 6=24$ and whose sum is equal to 25.
Such two numbers are 24 and 1 as $24+1=25$ and $24\times 1=24$.
Now, factoring the equation:
$\Rightarrow 4{{y}^{2}}+24y+y+6=0$
Taking common, we get:
$\begin{align}
  & \Rightarrow 4y\left( y+6 \right)+1\left( y+6 \right)=0 \\
 & \Rightarrow \left( y+6 \right)\left( 4y+1 \right)=0 \\
\end{align}$
Hence, $y+6=0$ or $4y+1=0$
$\Rightarrow y=-6$ or $y=-\dfrac{1}{4}$
Therefore, the roots of the equation are $y=-6,-\dfrac{1}{4}$.

Note: We must be careful enough to find the right common factors of the multiple terms given. We must also be careful enough while grouping the terms in order to avoid mistakes. One possible mistake that could be made while transposing terms that -6 could have been written as 6 which would produce incorrect roots of the equation.