Question

How do you factor $4{{m}^{4}}-37{{m}^{2}}+9$?

Answer 1

Hint: For answering this question we will use factorization. Factorization is the process of deriving factors of a number which divides the given number evenly. For this question we will split the constant term and we will use the sum product pattern of splitting and then we will simplify the equation using the \[\Rightarrow {{x}^{2}}-{{a}^{2}}=\left( x+a \right)\left( x-a \right)\] formula and simplify until we get the solution.

Complete step-by-step solution:
Now considering from the question we have an expression $4{{m}^{4}}-37{{m}^{2}}+9$ for which we need to derive the factors.
We can factor the $4{{m}^{4}}-37{{m}^{2}}+9$ by below method:
Given equation is in the form of \[a{{x}^{4}}+b{{x}^{2}}+c=0\].
First, we have to divide the coefficient of term in the equation which is $-37$ into the sum of the two numbers and must make sure that the product of the two numbers must be equal to the product of the coefficient of ${{m}^{4}}$ and constant that is product of $4$and $9$.
Now, the product can be split into the product of the two numbers as $-36\times -1$.
here we have to take the splitting of$-37$ as the sum of $-36$ and $-1$ which is equal to the coefficient of ${{m}^{2}}$.
Their sum is also equal to \[10\]which is equal to the coefficient of \[{{x}^{2}}\].
So, the given question can be factored as follows.
$\Rightarrow 4{{m}^{4}}-37{{m}^{2}}+9$
$\Rightarrow 4{{m}^{4}}-36{{m}^{2}}-{{m}^{2}}+9$
Here, we will take $4{{m}^{2}}$common in the first two terms and proceed further.
$\Rightarrow 4{{m}^{2}}\left( {{m}^{2}}-9 \right)-\left( {{m}^{2}}-9 \right)$
$\Rightarrow \left( 4{{m}^{2}}-1 \right)\left( {{m}^{2}}-9 \right)$
Here, we will use the formula \[\Rightarrow {{x}^{2}}-{{a}^{2}}=\left( x+a \right)\left( x-a \right)\] and proceed the further calculation.
$\Rightarrow \left( 2m-1 \right)\left( 2m+1 \right)\left( m-3 \right)\left( m+3 \right)$
Therefore, the factors will be$\Rightarrow \left( 2m-1 \right)\left( 2m+1 \right)\left( m-3 \right)\left( m+3 \right)$.

Note: During answering questions of this type, we should be sure with our calculations. Let us assume that ${{m}^{2}}=x$ so the equation will become as $\Rightarrow 4{{x}^{2}}-37x+9$ answering this question we can also use the formulae for obtaining the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ given as $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ then if the two solutions are $p,q$ then the factors will be $\left( x-p \right)\left( x-q \right)$ . For $\Rightarrow 4{{x}^{2}}-37x+9$ the roots are $\dfrac{37\pm \sqrt{1369-4\left( 9\times 4 \right)}}{2\times 4}=\dfrac{37\pm 35}{8}=\dfrac{72}{8},\dfrac{2}{8}=9,\dfrac{1}{4}$ then the factors will be $x-\dfrac{1}{4}$ and \[x-9\].
Later, we will replace ours \[{{m}^{2}}=x\]. So, the solution will be $\Rightarrow \left( {{m}^{2}}-\dfrac{1}{4} \right)\left( {{m}^{2}}-9 \right)$ and for further simplification we use \[\Rightarrow {{x}^{2}}-{{a}^{2}}=\left( x+a \right)\left( x-a \right)\] formula and get our final solution as follows $\Rightarrow \left( 2m-1 \right)\left( 2m+1 \right)\left( m-3 \right)\left( m+3 \right)$. Therefore, the factors will be $\Rightarrow \left( 2m-1 \right)\left( 2m+1 \right)\left( m-3 \right)\left( m+3 \right)$.