# 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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Answer

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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.

Complete step-by-step answer:

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 \[-1,-2,-3\].

Note: We observe that the equation given to us is a polynomial equation. Polynomial is an expression consisting of variables and coefficients that involves only the operations of addition, subtraction, multiplication or division, and non-negative integer exponents of variables. Degree of a polynomial is the value of the highest power of degrees of its individual term. We observe that the polynomial given to us is of degree 3. There are multiple ways to solve a polynomial equation, like completing the square and factoring the polynomial by splitting the intermediate terms. We have solved this question using the factorization method by splitting the intermediate terms. It’s necessary to keep in mind that we have to consider only integral solutions of the given equation. All other solutions will give an incorrect answer.

Complete step-by-step answer:

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 \[-1,-2,-3\].

Note: We observe that the equation given to us is a polynomial equation. Polynomial is an expression consisting of variables and coefficients that involves only the operations of addition, subtraction, multiplication or division, and non-negative integer exponents of variables. Degree of a polynomial is the value of the highest power of degrees of its individual term. We observe that the polynomial given to us is of degree 3. There are multiple ways to solve a polynomial equation, like completing the square and factoring the polynomial by splitting the intermediate terms. We have solved this question using the factorization method by splitting the intermediate terms. It’s necessary to keep in mind that we have to consider only integral solutions of the given equation. All other solutions will give an incorrect answer.

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