Courses
Courses for Kids
Free study material
Offline Centres
More
Store

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

Last updated date: 22nd Jul 2024
Total views: 451.5k
Views today: 5.51k
Verified
451.5k+ views
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 $-1,-2,-3$.