Question

How do you factor $45{{r}^{4}}+12r-30{{r}^{3}}-18{{r}^{2}}$ ?

Answer 1

Hint: First rearrange the terms in descending order of the exponentials. Now since $r$ is common in all the terms, take one $r$ common. Now take-out common terms from the first two terms and last two terms. Then write them together as the product of sums form. Now among the factorized terms, find the expression which can be further factorized and then write that also in the product of sums form. Now write all of them together and represent them as the factors of the given expression.

Complete step by step solution:
The given polynomial which must be factored is $45{{r}^{4}}+12r-30{{r}^{3}}-18{{r}^{2}}$
Now let us rearrange the terms in descending order of exponentials.
After arranging we get,
$\Rightarrow 45{{r}^{4}}-30{{r}^{3}}-18{{r}^{2}}+12r$
Now let us take one $r$ common out of all the terms.
After taking common we get,
$\Rightarrow r\left( 45{{r}^{3}}-30{{r}^{2}}-18r+12 \right)$
Now let us consider the coefficients in front of the variables.
All the coefficients have a common factor $3$ .
Let us take that also out.
After taking the common factor out we get,
$\Rightarrow 3r\left( 15{{r}^{3}}-10{{r}^{2}}-6r+4 \right)$
Now consider the polynomial $\left( 15{{r}^{3}}-10{{r}^{2}}-6r+4 \right)$
The polynomial is of degree $3$
Now, firstly let us take the common terms out of the first two terms.
$\Rightarrow 5{{r}^{2}}\left( 3r-2 \right)-6r+4$
Now secondly take the common terms out of the last two terms.
$\Rightarrow 5{{r}^{2}}\left( 3r-2 \right)-2\left( 3r+2 \right)$
Now on writing it in the form of the product of sums, also known as factoring,
$\Rightarrow \left( 5{{r}^{2}}-2 \right)\left( 3r+2 \right)$
Now we can see that there is still another polynomial left that can be factored further.
We can write $5{{r}^{2}}-2$ as ${{\left( \sqrt{5}r \right)}^{2}}-{{\left( \sqrt{2} \right)}^{2}}$
And since ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$
We can factorize it as,
$\Rightarrow \left( \sqrt{5}r+\sqrt{2} \right)\left( \sqrt{5}r-\sqrt{2} \right)$
Now writing it all together we get,
$\Rightarrow \left( 3r \right)\left( \sqrt{5}r+\sqrt{2} \right)\left( \sqrt{5}r-\sqrt{2} \right)\left( 3r+2 \right)$
Hence the factors for the polynomial $45{{r}^{4}}+12r-30{{r}^{3}}-18{{r}^{2}}$ are$\left( 3r \right)\left( \sqrt{5}r+\sqrt{2} \right)\left( \sqrt{5}r-\sqrt{2} \right)\left( 3r+2 \right)$

Note: Whenever there is a polynomial that is to be solved, the solution contains the roots of the expression. The number of roots is decided by the degree of the polynomial. The process of factorization is reverse multiplication. You can always cross-check your answer by placing the value of $x$ (factors of the expression) back in the equation. If you get LHS = RHS then your answer is correct.