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Hint: Type of question is based on the quadratic equation. As quadratic equation \[a{{x}^{2}}+2bx+c=0\]which can also be written as $ k(x-\alpha )(x-\beta ) $ where $ \alpha ,\beta $ are the roots of the quadratic equation, and K is the constant. Here, $ (x-\alpha ) $ and $ (x-\beta ) $ are known as the factor of the quadratic equation. We can easily find the quadratic factors of quadratic through the method of finding the roots, as if we know roots then we can easily interpret its factors. As we know that we can find the roots of quadratic through the splitting middle term method. The method says that the quadratic equation \[a{{x}^{2}}+2bx+c=0\] can be written in the format of\[{{x}^{2}}-(\alpha +\beta )x+\alpha \beta =0\], where $ \alpha ,\beta $ are the roots of the quadratic equation. When comparing the quadratic equation and spilling middle term equation we will get \[(\alpha +\beta )=\dfrac{-2b}{a}\]and $ \alpha \beta =\dfrac{c}{a} $ . We can get the $ \alpha ,\beta $ through the relation of $ a,b,c $ and $ \alpha ,\beta $ .
Complete step by step answer:
Moving further, with the question we had i.e. $ {{x}^{2}}-3x-10 $ of which we need to find out the factor. So, if we can find out the roots then we can easily write the factors also. So to find the roots we can go with the splitting middle term method.
By splitting middle term method $ {{x}^{2}}-3x-10 $ , $ -10 $ can be written as;
$ {{x}^{2}}-(3)x+(-10) $
Comparing it with\[{{x}^{2}}-(\alpha +\beta )x+\alpha \beta =0\],
Through which we can compute that
\[(\alpha +\beta )=3\] and $ \alpha \beta =-10 $ ;
So, $ \alpha \beta =-10 $ can be written as
$ (-5)\times 2 $ or $ 5\times (-2) $
$ \alpha =-5,\beta =2 $ or $ \alpha =5,\beta =-2 $ equation (i)
And\[(\alpha +\beta )=3\], it can be written as $ 5+(-2) $ , interpreting from above equation (i)
So, $ \alpha =5,\beta =-2 $ , interpreting from above calculation.
So, quadratic equation $ {{x}^{2}}-3x-10 $ can be written as $ {{x}^{2}}-[5+(-2)]x+[5(-2)]=0 $
Written through the method of splitting the middle term method, in which we got $ \alpha =5,\beta =-2 $ .
Which can further be written as $ [x-5][x-(-2)]=0 $ , derived from the $ k(x-\alpha )(x-\beta ) $ form.
From here we can say that \[[x-5]\]and $ [x-(-2)] $ are the factors of quadratic equation $ {{x}^{2}}-3x-10 $ .
Hence, \[[x-5]\]and $ [x+2] $ are the factors of a quadratic equation.
So, the correct answer is “Option C”.
Note: Keep in mind that for the factor form of equation i.e. $ k(x-\alpha )(x-\beta ) $ , x should have 1 as a coefficient. Moreover, in the general form of quadratic equation i.e. \[a{{x}^{2}}+2bx+c=0\], \[\dfrac{-2b}{a}\] is also called the sum of roots and $ \dfrac{c}{a} $ is called the product of roots.
Complete step by step answer:
Moving further, with the question we had i.e. $ {{x}^{2}}-3x-10 $ of which we need to find out the factor. So, if we can find out the roots then we can easily write the factors also. So to find the roots we can go with the splitting middle term method.
By splitting middle term method $ {{x}^{2}}-3x-10 $ , $ -10 $ can be written as;
$ {{x}^{2}}-(3)x+(-10) $
Comparing it with\[{{x}^{2}}-(\alpha +\beta )x+\alpha \beta =0\],
Through which we can compute that
\[(\alpha +\beta )=3\] and $ \alpha \beta =-10 $ ;
So, $ \alpha \beta =-10 $ can be written as
$ (-5)\times 2 $ or $ 5\times (-2) $
$ \alpha =-5,\beta =2 $ or $ \alpha =5,\beta =-2 $ equation (i)
And\[(\alpha +\beta )=3\], it can be written as $ 5+(-2) $ , interpreting from above equation (i)
So, $ \alpha =5,\beta =-2 $ , interpreting from above calculation.
So, quadratic equation $ {{x}^{2}}-3x-10 $ can be written as $ {{x}^{2}}-[5+(-2)]x+[5(-2)]=0 $
Written through the method of splitting the middle term method, in which we got $ \alpha =5,\beta =-2 $ .
Which can further be written as $ [x-5][x-(-2)]=0 $ , derived from the $ k(x-\alpha )(x-\beta ) $ form.
From here we can say that \[[x-5]\]and $ [x-(-2)] $ are the factors of quadratic equation $ {{x}^{2}}-3x-10 $ .
Hence, \[[x-5]\]and $ [x+2] $ are the factors of a quadratic equation.
So, the correct answer is “Option C”.
Note: Keep in mind that for the factor form of equation i.e. $ k(x-\alpha )(x-\beta ) $ , x should have 1 as a coefficient. Moreover, in the general form of quadratic equation i.e. \[a{{x}^{2}}+2bx+c=0\], \[\dfrac{-2b}{a}\] is also called the sum of roots and $ \dfrac{c}{a} $ is called the product of roots.
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