Question

Factorize ${{x}^{3}}-7x+6$

Answer 1

Hint:
Here we need to factorize the cubic equation. For that, we will find one random value which will satisfy the cubic equation. From there, we will get one factor of the cubic equation and then we will take that factor common from the equation. From there, we will get the quadratic equation also after taking that factor common. Then we will again factorize that quadratic equation also to get the complete factorization of the cubic equation.

Complete step by step solution:
The given cubic equation is
$\Rightarrow p\left( x \right)={{x}^{3}}-7x+6$
Now, we will substitute the value 1 in place of the variable $x$.
$\Rightarrow p\left( x \right)={{1}^{3}}-7\times 1+6$
On simplifying the terms, we get
$\Rightarrow p\left( x \right)=1-7+6$
On adding and subtracting the numbers, we get
$\Rightarrow p\left( x \right)=0$
We can see that the 1 is one of the roots of the given cubic polynomial.
Thus, we can say that $\left( x-1 \right)$ is the factor of the given cubic polynomial.
Now, we will take the common factor from the given cubic polynomial.
$\Rightarrow p\left( x \right)=\left( x-1 \right)\left( {{x}^{2}}+x-6 \right)$
Now, we will factorize the other factor of the given cubic polynomial which is a quadratic polynomial i.e. we will factorize $\left( {{x}^{2}}+x-6 \right)$ now.
First, we will split the middle term of the quadratic equation ${{x}^{2}}+x-6$ 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 $=-6$
The factors of 6 are 2 and 3. Therefore, we will split the middle term as;
$\Rightarrow {{x}^{2}}+x-6={{x}^{2}}+3x-2x+6$
Now, we will factorize the first two terms and last two terms separately.
$=x\left( x+3 \right)-2\left( x+3 \right)$
We can see that both new terms have a common factor.
Therefore,
$\Rightarrow 12{{x}^{2}}-7x+1=\left( x-2 \right)\left( x+3 \right)$
Hence, the factorization of ${{x}^{2}}+x-6$ is $\left( x-2 \right)\left( x+3 \right)$.

Hence, the factorization of given polynomial is $\Rightarrow p\left( x \right)=\left( x-1 \right)\left( x-2 \right)\left( x+3 \right)$

Note:
Remember that the number of roots of the polynomial is always equal to highest power in the polynomial. The number of roots of a quadratic polynomial is 2 because the highest power in a quadratic polynomial is 2 and similarly the number of roots of a cubic polynomial is 3 because the highest power in a cubic polynomial is 3.