Question

Factorise the polynomial ${{x}^{2}}+x-6$ ?

Answer 1

We need to factorise the given quadratic expression. We start to solve the given question by finding the roots of the quadratic equation using the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . Then, we write down the factors to get the desired result.

Complete step-by-step solution:
We are given a quadratic expression and are required to factorise the same. We will be solving the given question using the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .
The Quadratic equations are the polynomials with degree two. The quadratic equation will always have two roots. The roots may be real or imaginary.
The quadratic equation in standard form is given as follows,
$\Rightarrow f\left( x \right)=a{{x}^{2}}+bx+c$
Here,
a is the coefficient of ${{x}^{2}}$
b is the coefficient of $x$
c is the constant term
The roots of the quadratic equation $f\left( x \right)=a{{x}^{2}}+bx+c$ are found out using the formula,
$\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
According to our question,
We need to factorise the quadratic equation ${{x}^{2}}+x-6$
Comparing the quadratic equation ${{x}^{2}}+x-6$ with the standard form $a{{x}^{2}}+bx+c$ , we get,
a = 1;
b = 1;
c = -6.
We need to find the roots of the quadratic equation ${{x}^{2}}+x-6$
The roots of the quadratic equation ${{x}^{2}}+x-6$ are found out using the formula,
$\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Substituting the values of a, b, c in the above formula, we get,
$\Rightarrow x=\dfrac{-1\pm \sqrt{{{1}^{2}}-\left( 4\times 1\times -6 \right)}}{2\times 1}$
Simplifying the above equation, we get,
$\Rightarrow x=\dfrac{-1\pm \sqrt{1-\left( -24 \right)}}{2\times 1}$
Let us evaluate it further.
$\Rightarrow x=\dfrac{-1\pm \sqrt{1+24}}{2\times 1}$
$\Rightarrow x=\dfrac{-1\pm \sqrt{25}}{2\times 1}$
We know that the value of $\sqrt{25}$ is equal to 5. Substituting the same, we get,
$\Rightarrow x=\dfrac{-1\pm 5}{2}$
From the above,
$\Rightarrow x=\dfrac{-1+5}{2}\text{ or }x=\dfrac{-1-5}{2}$
Simplifying the above equations, we get,
$\Rightarrow x=\dfrac{4}{2}\text{ or }x=\dfrac{-6}{2}$
Canceling the common factors, we get,
$\therefore x=2\text{ or }x=-3$
If p, q are the roots of the quadratic equation $f\left( x \right)=a{{x}^{2}}+bx+c$ then the factors of the equation are given by $\left( x-p \right)\left( x-q \right)$
Following the same, we get,
$\Rightarrow {{x}^{2}}+x-6=\left( x-2 \right)\left( x-\left( -3 \right) \right)$
Simplifying the above equation, we get,
$\therefore {{x}^{2}}+x-6=\left( x-2 \right)\left( x+3 \right)$
Hence, the quadratic equation ${{x}^{2}}+x-6$ can be factored as $\left( x-2 \right)\left( x+3 \right)$.

Note: The result of the given question can be cross-checked using the equation ${{x}^{2}}+x-6=\left( x-2 \right)\left( x+3 \right)$ as follows,
The value of the LHS must be equal to RHS for any value of x. Let us suppose consider x = 3
LHS:
$\Rightarrow {{x}^{2}}+x-6$
Substituting the value of x = 3, we get,
$\Rightarrow {{3}^{2}}+3-6$
Simplifying the above expression, we get,
$\Rightarrow 9-3$
$\Rightarrow 6$
RHS:
$\Rightarrow \left( x-2 \right)\left( x+3 \right)$
Substituting the value of x = 3, we get,
$\Rightarrow \left( 3-2 \right)\left( 3+3 \right)$
Simplifying the above expression, we get,
$\Rightarrow \left( 1 \right)\left( 6 \right)$
$\Rightarrow 6$
LHS = RHS. The result attained is correct.