
How do you determine whether $x+2$ is a factor of the polynomial $2{{x}^{3}}+2{{x}^{2}}-x-6$
Answer
547.2k+ views
Hint: From the question we have been asked to check whether $x+2$ is a factor of the polynomial $2{{x}^{3}}+2{{x}^{2}}-x-6$ or not. So, here in this question we will use the factor theorem that is to determine whether $ax+b$ is a factor to a polynomial $p\left( x \right)$. We check \[p\left( -\dfrac{b}{a} \right)\], if it is equal to zero then it is factors to the polynomial.
Complete step-by-step solution:
Firstly, let us assume that a given polynomial is as follows.
$\Rightarrow p\left( x \right)=2{{x}^{3}}+2{{x}^{2}}-x-6$
So, here to determine whether $x+2$ is a factor of the polynomial or not we use the factor theorem mentioned above.
Now, we will find the \[p\left( -\dfrac{b}{a} \right)\] to be the polynomial. Where \[a=1,b=2\] when we compare the given $x+2$ with standard form $ax+b$
So, here the \[p\left( -\dfrac{b}{a} \right)\] will be \[p\left( -2 \right)\].
So, now we will find the \[p\left( -2 \right)\] by substituting this value in the polynomial $2{{x}^{3}}+2{{x}^{2}}-x-6$.
So, after substituting we get the equation as follows.
$\Rightarrow p\left( x \right)=2{{x}^{3}}+2{{x}^{2}}-x-6$
$\Rightarrow p\left( -2 \right)=2{{\left( -2 \right)}^{3}}+2{{\left( -2 \right)}^{2}}-\left( -2 \right)-6$
$\Rightarrow p\left( -2 \right)=-16+8+2-6$.
$\Rightarrow p\left( -2 \right)=-12$
Here we can clearly see that $p\left( -2 \right)=-12\ne 0$.
Here we after substituting the -2 in the polynomial we did not get 0. So, according to the factor theorem we can say that the given expression is not a factor to the polynomial.
Therefore, from this we can say that for the polynomial $2{{x}^{3}}+2{{x}^{2}}-x-6$ the expression $x+2$ is not a factor.
Note: Students must be very careful in doing the calculations. Students must have good knowledge in the concept of factor theorem and its applications.
Here we should not make mistakes like for example if we find \[p\left( 2 \right)\] instead of \[p\left( -2 \right)\] our solution according to the factor theorem will become a wrong one. So, we must be careful in this aspect of the problem.
Complete step-by-step solution:
Firstly, let us assume that a given polynomial is as follows.
$\Rightarrow p\left( x \right)=2{{x}^{3}}+2{{x}^{2}}-x-6$
So, here to determine whether $x+2$ is a factor of the polynomial or not we use the factor theorem mentioned above.
Now, we will find the \[p\left( -\dfrac{b}{a} \right)\] to be the polynomial. Where \[a=1,b=2\] when we compare the given $x+2$ with standard form $ax+b$
So, here the \[p\left( -\dfrac{b}{a} \right)\] will be \[p\left( -2 \right)\].
So, now we will find the \[p\left( -2 \right)\] by substituting this value in the polynomial $2{{x}^{3}}+2{{x}^{2}}-x-6$.
So, after substituting we get the equation as follows.
$\Rightarrow p\left( x \right)=2{{x}^{3}}+2{{x}^{2}}-x-6$
$\Rightarrow p\left( -2 \right)=2{{\left( -2 \right)}^{3}}+2{{\left( -2 \right)}^{2}}-\left( -2 \right)-6$
$\Rightarrow p\left( -2 \right)=-16+8+2-6$.
$\Rightarrow p\left( -2 \right)=-12$
Here we can clearly see that $p\left( -2 \right)=-12\ne 0$.
Here we after substituting the -2 in the polynomial we did not get 0. So, according to the factor theorem we can say that the given expression is not a factor to the polynomial.
Therefore, from this we can say that for the polynomial $2{{x}^{3}}+2{{x}^{2}}-x-6$ the expression $x+2$ is not a factor.
Note: Students must be very careful in doing the calculations. Students must have good knowledge in the concept of factor theorem and its applications.
Here we should not make mistakes like for example if we find \[p\left( 2 \right)\] instead of \[p\left( -2 \right)\] our solution according to the factor theorem will become a wrong one. So, we must be careful in this aspect of the problem.
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