Question

How do you factor \[{{x}^{2}}-8x-33\]?

Answer 1

Hint: This question is from the topic of quadratic equation. We will find the factor of the equation \[{{x}^{2}}-8x-33\] using Sridharacharya’s rule. In solving this question, we will first understand what Sridharacharya’s rule is. After that, we will find out the roots of the equation. After solving further processes, we will find out the factor of the equation \[{{x}^{2}}-8x-33\].

Complete step by step solution:Let us solve this question.
In this question, we have asked to factor the term \[{{x}^{2}}-8x-33\].
Let us first understand about Sridhracharya’s rule.
This rule says that if we have a general equation like \[a{{x}^{2}}+bx+c\], then the roots or zeros of equation will be \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] that is \[\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}\] and \[\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}\].
So, from the term \[{{x}^{2}}-8x-33\], we can write the zeros of x as
\[x=\dfrac{-\left( -8 \right)\pm \sqrt{{{\left( -8 \right)}^{2}}-4\times 1\times \left( -33 \right)}}{2\times 1}\]
The above can also be written as
\[x=\dfrac{8\pm \sqrt{64+132}}{2}\]
The above can also be written as
\[x=\dfrac{8\pm \sqrt{196}}{2}\]
As we that square of 14 is 196 or we can say square root of 196 is 14, so we can write the above equation as
\[\Rightarrow x=\dfrac{8\pm 14}{2}\]
Now, after dividing 2, we get
\[\Rightarrow x=4\pm 7\]
Hence, we get that the values of x are \[4+7\] and \[4-7\]
Or, we can say the values of x are 11 and -3.
As we know that, if \[{{x}_{1}}\] is a root of the equation any equation of variable x, then \[\left( x-{{x}_{1}} \right)\] is a factor of the same equation.
So, we can say that \[\left( x-11 \right)\] and \[\left( x+3 \right)\] are factors of the equation \[{{x}^{2}}-8x-33\].
Or, we can say that the factor of equation \[{{x}^{2}}-8x-33\] is \[\left( x-11 \right)\left( x+3 \right)\].

Note: We should have a better knowledge in the topic of algebra. We should know that according to Sridharacharya’s rule if we have an equation \[a{{x}^{2}}+bx+c\], then the zeros of this equation will be \[\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}\] and \[\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}\] & the factor of the same equation will be \[\left( x-\left[ \dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a} \right] \right)\left( x-\left[ \dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a} \right] \right)\]
We can solve this question by alternate method.
The term from which we have to factor is
\[{{x}^{2}}-8x-33\]
The above can also be written as