Question

How do you factor $30{x^3} = 21{x^2} + 135{x^2}$ ?

Answer 1

Hint: The question belongs to the simplification of the polynomial having degree three in one variable. In this type of problem, we will look for a common factor in the given polynomial. We will factor a polynomial having degree two. To factor the polynomial, we will multiply the constant term in the polynomial with the coefficients of ${x^2}$, and then find a factor, which in addition or subtraction will give the middle term of the polynomial. We will again look for a common factor in the polynomial. Then we will equate the factors equal to zero, to find the solution of the polynomial. We can also solve the given polynomial in another way. We substitute the different values of $x$ which satisfies the given polynomial. If the value gets satisfied then it will be the factor of the polynomial. Similarly we do until the factorization of polynomials is completely done. The common factor will also be the factor of the given polynomial.

Complete step by step solution:
Step: 1 the given polynomial is,
$30{x^3} = 21{x^2} + 135x$
Take all the terms of the equation one side.
$30{x^3} - 21{x^2} - 135x = 0$
Now look for the common factor in the polynomial. Here we will take $3{x^2}$ as common.
Therefore,
$ \Rightarrow 30{x^3} - 21{x^2} - 135x = 3{x^2}\left( {10{x^2} - 7x - 45} \right)$
Solve the quadratic equation to find the factor of the polynomial. First multiply the constant term with the coefficient of ${x^2}$ then find a factor of the term in the quadratic equation so that on addition or subtraction the factors, it will give the middle term of the quadratic equation.
$ \Rightarrow 10 \times 45 = 450$
$ \Rightarrow 450 = 18 \times 25$
Step: 2 consider the factor $18 \times 25$ , which in addition will give you the middle term of the quadratic equation.
$ \Rightarrow 3{x^2}\left( {10{x^2} - 7x - 45} \right) = 3{x^2}\left( {10{x^2} + 18x - 25x - 45} \right)$
$ \Rightarrow 3{x^2}\left( {10{x^2} + 18x - 25x - 45} \right) = 3{x^2}\left\{ {2x\left( {5x + 9} \right) - 5\left( {5x + 9} \right)} \right\}$
$ \Rightarrow 3{x^2}\left\{ {2x\left( {5x + 9} \right) - 5\left( {5x + 9} \right)} \right\} = 3{x^2}\left( {2x - 5} \right)\left( {5x + 9} \right)$
 Step: 3 equate the factors equal to zero of the polynomial.
$ \Rightarrow {x^2}\left( {2x - 5} \right)\left( {5x + 9} \right) = 0$
$
   \Rightarrow 3{x^2} = 0 \\
   \Rightarrow x = 0 \\
 $
Therefore,
$
   \Rightarrow \left( {5x + 9} \right) = 0 \\
   \Rightarrow x = \dfrac{{ - 9}}{5} \\
 $
And,
$
   \Rightarrow \left( {2x - 5} \right) = 0 \\
   \Rightarrow x = \dfrac{5}{2} \\
 $

Final Answer:
Therefore the factors of the given polynomial are $x = 0,\dfrac{5}{2},\dfrac{{ - 9}}{5}$.

Note:
Simplify the given polynomial by taking common into the equation. Solve the quadratic equation by using the middle term concept, first multiply the constant term of the quadratic equation with the coefficient of ${x^2}$ and find the factor of the number so that on addition or subtraction which gives the middle term of the equation.