Question

How do you factor $14{x^2}{y^3} + 5x{y^3} - 9{y^3}$ ?

Answer 1

Hint: In this question, we need to find the factors of the given polynomial. Firstly, we will factor out the common term which is ${y^3}$ from the given equation. Then we obtain a quadratic equation which we solve using the factoring method. This polynomial will be of the form $a{x^2} + bx + c$. So we rewrite the middle term as a sum of terms whose product is $a \cdot c$ and whose sum is b. Therefore, substitute the values of a, b, c and solve the given problem by splitting the middle term and obtain the required factors.

Complete step by step solution:
Given a polynomial of the form $14{x^2}{y^3} + 5x{y^3} - 9{y^3}$ …… (1)
We are asked to factorize the polynomial given in the equation (1).
If we carefully observe the polynomial, the term ${y^3}$ is common throughout the polynomial.
So factoring out the common term ${y^3}$ in the equation (1), we get,
$ \Rightarrow 14{x^2}{y^3} + 5x{y^3} - 9{y^3} = {y^3}(14{x^2} + 5x - 9)$ …… (2)
Note that the polynomial inside the parenthesis is a quadratic polynomial.
Now consider the quadratic polynomial $14{x^2} + 5x - 9$.
Consider an equation of the form $a{x^2} + bx + c$, where a, b, c are any real numbers.
We rewrite the middle term as a sum of two terms in such a way that their product is $a \cdot c$ and their sum is b.
In the given equation we have $a = 14$, $b = 5$ and $c = - 9$.
We split the middle term $5x$ as,$5x = 14x - 9x$.
Note that their product is,
 $a \cdot c = 14 \times ( - 9)$
$ \Rightarrow a \cdot c = - 126$
Note that their sum is,
$b = 5$
$ \Rightarrow 5 = 14 - 9$.
Hence the equation $14{x^2} + 5x - 9$ can be written as,
$ \Rightarrow 14{x^2} + 14x - 9x - 9$
Factor out the greatest common factor from each group, we get,
$ \Rightarrow 14x(x + 1) - 9(x + 1)$
Now factor the polynomial by factoring out the greatest common factor $x + 1$, we get,
$ \Rightarrow (x + 1)(14x - 9)$
Hence the factorization of the equation $14{x^2} + 5x - 9$ is $(x + 1)(14x - 9)$.
Substituting this in the equation (2), we get,
$ \Rightarrow 14{x^2}{y^3} + 5x{y^3} - 9{y^3} = {y^3}(x + 1)(14x - 9)$.

Hence the factorization of the polynomial $14{x^2}{y^3} + 5x{y^3} - 9{y^3}$ gives us the factors as, ${y^3}(x + 1)(14x - 9)$.

Note: If any term in the polynomial is common, we need to factor it out, so that it will be easier to find the solution for the problem. When we multiply the factors obtained with each other, then we must obtain the original polynomial given. If we fail in obtaining it, then our factors are not correct.
Students must know how to apply factorization method to the quadratic polynomial.
If we have an equation of the form $a{x^2} + bx + c$, where a, b, c are any real numbers.
To find the factors, we rewrite the middle term as a sum of two terms in such a way that their product is $a \cdot c$ and their sum is b. So we obtain the required factors after simplification.