Question

Factorize:
A) $12{{x}^{2}}-7x+1$
B) $2{{x}^{2}}+7x+3$
C) $6{{x}^{2}}+5x-6$
D) $3{{x}^{2}}-x-4$

Answer 1

Hint:
Here we have to factorize the quadratic equations. For that, we will split the middle term into two terms, whose coefficients are the factors of the term obtained from the multiplication of the coefficient of $x{}^{2}$ and the constant term. Then we will factorize the first two terms and the last two terms separately. Then we will get the common factor from the new terms, after taking common factors, we will get the final factorization of the required quadratic equation.

Complete step by step solution:
A) $12{{x}^{2}}-7x+1$
First, we will split the middle term of the quadratic equation $12{{x}^{2}}-7x+1$into two new terms. We have to split the terms in such a way that the coefficients of the two terms are the factors of the term obtained from the multiplication of the coefficient of $x{}^{2}$ and the constant term.
Multiplication of the coefficient of $x{}^{2}$ and the constant term $=12$
The factors of 12 are 4 and 3. Therefore, we will split the middle term as;
$\Rightarrow 12{{x}^{2}}-7x+1=12{{x}^{2}}-4x-3x+1$
Now, we will factorize the first two terms and last two terms separately.
$\Rightarrow 12{{x}^{2}}-7x+1=4x\left( 3x-1 \right)-\left( 3x-1 \right)$
We can see that both new terms have a common factor.
Therefore,
$\Rightarrow 12{{x}^{2}}-7x+1=\left( 4x-1 \right)\left( 3x-1 \right)$
Hence, the factorization of $12{{x}^{2}}-7x+1$ is $\left( {4x - 1} \right)\left( {3x - 1} \right)$.
B) $2{{x}^{2}}+7x+3$
First, we will split the middle term of the quadratic equation $2{{x}^{2}}+7x+3$into two new terms. We have to split the terms in such a way that the coefficients of the two terms are the factors of the term obtained from the multiplication of the coefficient of $x{}^{2}$ and the constant term.
Multiplication of the coefficient of $x{}^{2}$ and the constant term $=6$
The possible factors of 6 are 2, 3, 6 and 1. Therefore, we will split the middle term as;
$\Rightarrow 2{{x}^{2}}+7x+3=2{{x}^{2}}+6x+x+3$
Now, we will factorize the first two terms and last two terms separately.
$\Rightarrow 2{{x}^{2}}+7x+3=2x\left( x+3 \right)+\left( x+3 \right)$
We can see that both new terms have a common factor.
Therefore,
$\Rightarrow 2{{x}^{2}}+7x+3=\left( 2x+1 \right)\left( x+3 \right)$
Hence, the factorization of $2{{x}^{2}}+7x+3$ is $\left( 2x+1 \right)\left( x+3 \right)$.
C) $6{{x}^{2}}+5x-6$
First, we will split the middle term of the quadratic equation $6{{x}^{2}}+5x-6$into two new terms. We have to split the terms in such a way that the coefficients of the two terms are the factors of the term obtained from the multiplication of the coefficient of $x{}^{2}$ and the constant term.
Multiplication of the coefficient of $x{}^{2}$ and the constant term $=36$
The possible factors of 36 are 2, 3, 6, 9 and 4. Therefore, we will split the middle term as;
$\Rightarrow 6{{x}^{2}}+5x-6=6{{x}^{2}}+9x-4x-6$
Now, we will factorize the first two terms and last two terms separately.
$\Rightarrow 6{{x}^{2}}+5x-6=3x\left( 2x+3 \right)-2\left( 2x+3 \right)$
We can see that both new terms have a common factor.
Therefore,
$\Rightarrow 6{{x}^{2}}+5x-6=\left( 3x-2 \right)\left( 2x+3 \right)$
Hence, the factorization of $6{{x}^{2}}+5x-6$ is $\left( {3x - 2} \right)\left( {2x + 3} \right)$.
D) $3{{x}^{2}}-x-4$
First, we will split the middle term of the quadratic equation $3{{x}^{2}}-x-4$into two new terms. We have to split the terms in such a way that the coefficients of the two terms are the factors of the term obtained from the multiplication of the coefficient of $x{}^{2}$ and the constant term.
Multiplication of the coefficient of $x{}^{2}$ and the constant term $=12$
The possible factors of 36 are 2, 3, and 4. Therefore, we will split the middle term as;
$\Rightarrow 3{{x}^{2}}-x-4=3{{x}^{2}}-4x+3x-4$
Now, we will factorize the first two terms and last two terms separately.
$\Rightarrow 3{{x}^{2}}-x-4=x\left( 3x-4 \right)+\left( 3x-4 \right)$
We can see that both new terms have a common factor.
Therefore,
$\Rightarrow 3{{x}^{2}}-x-4=\left( x+1 \right)\left( 3x-4 \right)$

Hence, the factorization of $3{{x}^{2}}-x-4$ is $\left( x+1 \right)\left( 3x-4 \right)$.

Note:
An equation having its degree 2 is called Quadratic Equation. Example - ${{x}^{2}}+x+1=0$ ,
$2{{y}^{2}}+3y-5=0$ etc.
General Form of Quadratic Equation is;
$a{{x}^{2}}+b{{y}^{2}}+2hxy+2gx+2fy+c=0$.
This equation can represent a Pair of Straight Lines, Circle, Ellipse, Parabola and Hyperbola.