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# Find the integral roots of the polynomial $f\left( x \right)={{x}^{3}}+6{{x}^{2}}+11x+6$?

Last updated date: 18th Mar 2023
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Hint: To find the integral roots of the polynomial $f\left( x \right)={{x}^{3}}+6{{x}^{2}}+11x+6$, we will equate the polynomial to zero and factorize the polynomial by adding or subtracting some terms from the polynomial. This method is called factoring the polynomial by splitting the intermediate terms.

We have the polynomial $f\left( x \right)={{x}^{3}}+6{{x}^{2}}+11x+6$. We have to find the integral roots of the polynomial. So, we will equate the polynomial to zero and factorize it to get the roots of the polynomial.
Thus, we have $f\left( x \right)={{x}^{3}}+6{{x}^{2}}+11x+6=0$.
We can write the above expression as ${{x}^{3}}+{{x}^{2}}+5{{x}^{2}}+5x+6x+6=0$.
Taking out the common terms, we have ${{x}^{2}}\left( x+1 \right)+5x\left( x+1 \right)+6\left( x+1 \right)=0$.
Thus, we have $\left( x+1 \right)\left( {{x}^{2}}+5x+6 \right)=0$.
We can rewrite the above equation as $\left( x+1 \right)\left( {{x}^{2}}+2x+3x+6 \right)=0$.
Taking out the common terms, we have $\left( x+1 \right)\left\{ x\left( x+2 \right)+3\left( x+2 \right) \right\}=0$.
Thus, we have $\left( x+1 \right)\left( x+2 \right)\left( x+3 \right)=0$.
So, we get $x=-1,-2,-3$.
Hence, the integral roots of the polynomial $f\left( x \right)={{x}^{3}}+6{{x}^{2}}+11x+6$ are $-1,-2,-3$.