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

Last updated date: 18th Jul 2024
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Hint: The factor theorem is a special case of the polynomial remainder theorem. According to the factor theorem, a polynomial $f\left( x \right)$ has a factor $\left( {x - k} \right)$ if and only if $f\left( k \right) = 0$. We will first equate the divisor polynomial to zero so as to get the value of the variable to be put in the dividend polynomial in order to find the remainder. If the remainder comes out to be zero, we get our required result.

Complete step-by-step solution:
In the given question, we are provided with two functions $g\left( x \right)$ and $p\left( x \right)$. So, we have to show that the polynomial $g\left( x \right)$ is a factor of polynomial $p\left( x \right)$ using the factor theorem.
So, according to the factor theorem,
For $\left( {x - k} \right)$ to be a factor of $f\left( x \right)$, $f\left( k \right)$ has to be equal to zero.
So, we have the dividend as $p\left( x \right) = 69 + 11x - {x^2} + {x^3}$and divisor as $g\left( x \right) = x + 3$.
First equating the divisor $g\left( x \right)$ as zero, we get the value of x as,
$\Rightarrow g\left( x \right) = x + 3 = 0$
$\Rightarrow x = - 3$
So, for $g\left( x \right) = x + 3$ to be a factor of $p\left( x \right) = 69 + 11x - {x^2} + {x^3}$, the value of $p\left( { - 3} \right)$ should be equal to zero so that the remainder upon the division of the polynomials is zero.
So, we substitute the value of x as $\left( { - 3} \right)$ in the original dividend function $p\left( x \right) = 69 + 11x - {x^2} + {x^3}$. Hence, we get,
$\Rightarrow p\left( { - 3} \right) = 69 + 11\left( { - 3} \right) - {\left( { - 3} \right)^2} + {\left( { - 3} \right)^3}$
Now, evaluating the powers of $\left( { - 3} \right)$, we get,
$\Rightarrow p\left( { - 3} \right) = 69 - 33 - 9 + \left( { - 27} \right)$
Opening the brackets, we get,
$\Rightarrow p\left( { - 3} \right) = 69 - 33 - 9 - 27$
Adding up the terms with like signs, we get,
$\Rightarrow p\left( { - 3} \right) = 69 - 69$
Simplifying further, we get,
$\Rightarrow p\left( { - 3} \right) = 0$
So, the value of $p\left( { - 3} \right)$ equals zero.
Hence, $g\left( x \right)$ is a factor of $p\left( x \right)$.

Note: Factor theorem is a theorem that links factors and zeros of a polynomial. Substitution of a variable involves putting a certain value in place of the variable. That specified value may be a certain number or even any other variable. We must ensure accuracy in calculations and substitution of variables. We also must remember that odd powers of a negative number always yield negative results.