Question

How do you solve $2{x^2} - 13x + 15 = 0$ by factoring?

Answer 1

Hint:First find the product of first and last constant term of the given expression. Then, choose the factors of product value in such a way that addition or subtraction of those factors is the middle constant term. Then split the middle constant term or coefficient of $x$ in these factors and take common terms out in first terms and last two terms. Then again take common terms out of terms obtained. Then, equate both factors to $0$ and determine the value of $x$.

Formula used:
For factorising an algebraic expression of the type $a{x^2} + bx + c$, we find two factors $p$ and $q$ such that
$ac = pq$ and $p + q = b$

Complete step by step answer:
Given, $2{x^2} - 13x + 15 = 0$
We have to factor this quadratic equation.
To factor this quadratic equation, first we have to find the product of the first and last constant term of the expression.
Here, first constant term in $2{x^2} - 13x + 15 = 0$ is $2$, as it is the coefficient of ${x^2}$ and last constant term is $15$, as it is a constant value.
Now, we have to multiply the coefficient of ${x^2}$ with the constant value in $2{x^2} - 13x + 15 = 0$, i.e., multiply $2$ with $15$.
Multiplying $2$ and $15$, we get
$2 \times 15 = 30$
Now, we have to find the factors of $30$ in such a way that addition or subtraction of those factors is the middle constant term.
Middle constant term or coefficient of $x$ in $2{x^2} - 13x + 15 = 0$ is $ - 13$.
So, we have to find two factors of $30$, which on multiplying gives $30$ and in addition gives $ - 13$.
We can do this by determining all factors of $30$.
Factors of $30$ are $ \pm 1, \pm 2, \pm 3, \pm 5, \pm 6, \pm 10, \pm 15, \pm 30$.
Now among these values find two factors of $30$, which on multiplying gives $30$ and in addition gives $ - 13$.
After observing, we can see that
$\left( { - 3} \right) \times \left( { - 10} \right) = 30$ and $\left( { - 3} \right) + \left( { - 10} \right) = - 13$
So, these factors are suitable for factorising the given trinomial.
Now, the next step is to split the middle constant term or coefficient of $x$ in these factors.
That is, write $ - 13x$ as $ - 3x - 10x$ in $2{x^2} - 13x + 15 = 0$.
After writing $ - 13x$ as $ - 3x - 10x$ in $2{x^2} - 13x + 15 = 0$, we get
$ \Rightarrow 2{x^2} - 3x - 10x + 15 = 0$
Now, taking $x$ common in $\left( {2{x^2} - 3x} \right)$ and putting in above equation, we get
$ \Rightarrow x\left( {2x - 3} \right) - 10x + 15 = 0$
Now, taking $ - 5$ common in $\left( { - 10x + 15} \right)$ and putting in above equation, we get
$ \Rightarrow x\left( {2x - 3} \right) - 5\left( {2x - 3} \right) = 0$
Now, taking $\left( {2x - 3} \right)$common in $x\left( {2x - 3} \right) - 5\left( {2x - 3} \right)$ and putting in above equation, we get
$ \Rightarrow \left( {2x - 3} \right)\left( {x - 5} \right) = 0$
Now, equate both factors to $0$ and determine the value of $x$.
So, $2x - 3 = 0$ and $x - 5 = 0$
$ \Rightarrow x = \dfrac{3}{2}$ and $x = 5$

Therefore, $x = \dfrac{3}{2}$ and $x = 5$ are solutions of $2{x^2} - 13x + 15 = 0$.

Note: In above question, it should be noted that we took $ - 3$ and $ - 10$ as factors of $ - 30$, which on multiplying gives $ - 30$ and in addition gives $ - 13$. No, other factors will satisfy the condition. If we take wrong factors, then we will not be able to take common terms out in the next step. So, carefully select the numbers.