Question

How do you find all the zeros of $F\left( x \right)=9{{x}^{4}}-37{{x}^{2}}+4$ ?

Answer 1

Hint: We are trying to find the zeros of the term given in the problem. To start with, we will consider the value ${{x}^{2}}=u$ and simplify it like a middle term problem. Then, putting the value back and using the algebraic formula of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$, we will reach our desired result.

Complete step by step solution:
According to the question, we are trying to find the zeros of the equation, $F\left( x \right)=9{{x}^{4}}-37{{x}^{2}}+4$.
To make the solution and process a more simplified one, we start by considering, ${{x}^{2}}=u$
Thus, we have the equation as, $F\left( u \right)=9{{u}^{2}}-37u+4$
We will try to solve this using the middle term factor.
By multiplying the first and third coefficients, we are getting, $\left( 9\times 4 \right)=36$ .
Also, 36 can be written as, $36=36\times 1$ .
By adding them we get the second coefficient, 36 + 1 = 37.
Thus, now, $F\left( u \right)=9{{u}^{2}}-37u+4$can be written as,
$\Rightarrow 9{{u}^{2}}-37u+4=9{{u}^{2}}-36u-u+4$
Now, taking 9u common from the first two terms and -1 common from the last two terms, we are getting a similarity.
Doing that, we get, $9u\left( u-4 \right)-1\left( u-4 \right)$
Now, again, writing $u-4$ together, we have, $\left( 9u-1 \right)\left( u-4 \right)$
But, we also have, ${{x}^{2}}=u$.
Putting back the values, now, $\left( 9u-1 \right)\left( u-4 \right)=\left( 9{{x}^{2}}-1 \right)\left( {{x}^{2}}-4 \right)$
This also can be factorized.
$\left( 9{{x}^{2}}-1 \right)\left( {{x}^{2}}-4 \right)=\left( {{\left( 3x \right)}^{2}}-{{\left( 1 \right)}^{2}} \right)\left( {{\left( x \right)}^{2}}-{{\left( 2 \right)}^{2}} \right)$
From the known formula of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ , we are now getting,
${{\left( 3x \right)}^{2}}-{{\left( 1 \right)}^{2}}=\left( 3x+1 \right)\left( 3x-1 \right)$
And also,
${{\left( x \right)}^{2}}-{{\left( 2 \right)}^{2}}=\left( x+2 \right)\left( x-2 \right)$
Hence, the factor of our given equation goes like, $\left( 9{{x}^{2}}-1 \right)\left( {{x}^{2}}-4 \right)=\left( 3x+1 \right)\left( 3x-1 \right)\left( x+2 \right)\left( x-2 \right)$
Equaling this with the value zero, we will get each of them having a value zero if other values are non zero.
So, $3x+1=0$ gives us,
$\Rightarrow 3x=-1$
Dividing both sides by 3, we get,
$\Rightarrow x=-\dfrac{1}{3}$
And , $3x-1=0$ gives us,
$\Rightarrow 3x=1$
Dividing both sides by 3, we get,
$\Rightarrow x=\dfrac{1}{3}$
Also, $x+2=0$ gives us,
$\Rightarrow x=-2$
Also, $x-2=0$ gives us,
$\Rightarrow x=2$
So, the zeros of the given term is, $x=-\dfrac{1}{3},\dfrac{1}{3},2,-2$.

Note: A polynomial $F\left( x \right)=9{{x}^{4}}-37{{x}^{2}}+4$ can be written as a product of two or more polynomials of degree less than or equal to that of it. Each polynomial involved in the product will be a factor of it. Monomials can be factorized in the same way as integers, just by writing the monomial as the product of its constituent prime factors. In the case of monomials, these prime factors can be integers as well as other monomials which cannot be factored further.