# Using factor theorem show that \[g\left( x \right)\] is a factor of \[p\left( x \right)\]\[=69+11x-{{x}^{2}}+{{x}^{3}}\], \[g\left( x \right)=x+3\].

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**Hint:**In order to show that \[g\left( x \right)=x+3\] is a factor of \[p\left( x \right)\]\[=69+11x-{{x}^{2}}+{{x}^{3}}\], we must solve for \[g\left( x \right)\] at first. Then we are supposed to substitute this value of \[g\left( x \right)\] in \[p\left( x \right)\] and solve the polynomial. If the value of \[p\left( x \right)\] would be zero, then the given \[g\left( x \right)\] will be a factor of \[p\left( x \right)\].

**Complete step-by-step solution:**

Now let us learn about the factor theorem. The factor theorem is a theorem that links the factors and zeros of a polynomial. This is a special case of the remainder theorem. According to this theorem we say that \[x-a\] is a factor of \[f\left( x \right)\], if \[f\left( a \right)=0\]. This theorem is commonly used to find the roots of the polynomial. We can also find the factor of a polynomial by other methods such as the polynomial long division method and the synthetic division method.

Now let us find if \[g\left( x \right)=x+3\] is a factor of \[p\left( x \right)\]\[=69+11x-{{x}^{2}}+{{x}^{3}}\].

Firstly, we must be solving for \[g\left( x \right)=x+3\].

We get,

\[\begin{align}

& g\left( x \right)=x+3 \\

& \Rightarrow x+3=0 \\

& \Rightarrow x=-3 \\

\end{align}\]

We obtain the value of \[x\] as \[-3\].

Now we will be substituting this obtained value in \[p\left( x \right)\]\[=69+11x-{{x}^{2}}+{{x}^{3}}\].

\[\begin{align}

& \Rightarrow p\left( x \right)=69+11x-{{x}^{2}}+{{x}^{3}} \\

& \Rightarrow p\left( -3 \right)=69+11\left( -3 \right)-{{\left( -3 \right)}^{2}}+{{\left( -3 \right)}^{3}} \\

& \Rightarrow p\left( -3 \right)=69-33-9-27 \\

& \Rightarrow p\left( -3 \right)=0 \\

\end{align}\]

We see that we have obtained \[p\left( x \right)=0\].

**\[\therefore \] We can conclude that given \[g\left( x \right)=x+3\] is a factor of \[p\left( x \right)\]\[=69+11x-{{x}^{2}}+{{x}^{3}}\].**

**Note:**To apply the factor theorem, we must always have a note that for a polynomial \[f\left( x \right)\] the degree of the polynomial should be greater than or equal to one. The degree of the polynomial is nothing but the highest power of the term in the expression. The value which solves the expression or equation is known as polynomial value.

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