
How do you factor the polynomial $4{{x}^{5}}-9{{x}^{3}}$?
Answer
546.3k+ views
Hint: We are given a polynomial which has to be solved by the method of factoring the equation. We will group the common terms and form two equations in x-variable out of the given quadratic equation. After further equating each of the grouped equations to zero, we shall 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, $4{{x}^{5}}-9{{x}^{3}}$,
$\Rightarrow 4{{x}^{5}}-9{{x}^{3}}=0$
We shall take ${{x}^{3}}$ common and group the remaining terms.
Taking common, we get:
$\Rightarrow {{x}^{3}}\left( 4{{x}^{2}}-9 \right)=0$
Hence, ${{x}^{3}}=0$ or $4{{x}^{2}}-9=0$
For ${{x}^{3}}=0$,
When we take cube root the left hand side as well the right hand side of the equation, we get
$\Rightarrow \sqrt[3]{{{x}^{3}}}=\sqrt[3]{0}$
$\Rightarrow x=0$
For $4{{x}^{2}}-9=0$,
We shall first transpose 9 to the right hand side and then divide both sides of the equation by 4.
$\Rightarrow {{x}^{2}}=\dfrac{9}{4}$
Further, we will find the square root of the entire equation.
$\Rightarrow \sqrt{{{x}^{2}}}=\sqrt{\dfrac{9}{4}}$
We know that $4={{2}^{2}}$ or $\sqrt{4}=2$ and $9={{3}^{2}}$ or $\sqrt{9}=3$
$\Rightarrow x=\pm \dfrac{3}{2}$
Therefore, the roots of the equation are $x=0,\dfrac{3}{2},-\dfrac{3}{2}$.
Note: There must be 5 roots of the given equation as it is a 5-degree equation. However, we could calculate only 3 roots because the other two roots are repeating and are equal to 0. Also, while calculating the square root of any term, we must not forget to add both the positive as well as negative signs along with it.
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, $4{{x}^{5}}-9{{x}^{3}}$,
$\Rightarrow 4{{x}^{5}}-9{{x}^{3}}=0$
We shall take ${{x}^{3}}$ common and group the remaining terms.
Taking common, we get:
$\Rightarrow {{x}^{3}}\left( 4{{x}^{2}}-9 \right)=0$
Hence, ${{x}^{3}}=0$ or $4{{x}^{2}}-9=0$
For ${{x}^{3}}=0$,
When we take cube root the left hand side as well the right hand side of the equation, we get
$\Rightarrow \sqrt[3]{{{x}^{3}}}=\sqrt[3]{0}$
$\Rightarrow x=0$
For $4{{x}^{2}}-9=0$,
We shall first transpose 9 to the right hand side and then divide both sides of the equation by 4.
$\Rightarrow {{x}^{2}}=\dfrac{9}{4}$
Further, we will find the square root of the entire equation.
$\Rightarrow \sqrt{{{x}^{2}}}=\sqrt{\dfrac{9}{4}}$
We know that $4={{2}^{2}}$ or $\sqrt{4}=2$ and $9={{3}^{2}}$ or $\sqrt{9}=3$
$\Rightarrow x=\pm \dfrac{3}{2}$
Therefore, the roots of the equation are $x=0,\dfrac{3}{2},-\dfrac{3}{2}$.
Note: There must be 5 roots of the given equation as it is a 5-degree equation. However, we could calculate only 3 roots because the other two roots are repeating and are equal to 0. Also, while calculating the square root of any term, we must not forget to add both the positive as well as negative signs along with it.
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