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**Hint:**We have been given a polynomial expression $p\left( x \right)$ and also given another expression $x - a$ is a factor of the given polynomial. We have to determine the remaining factors of the given polynomial. According to the remainder theorem, if $x - a$ is a factor of $p\left( x \right)$ then $p\left( a \right) = 0$ . To find the other factors of the given polynomial, first, we divide the given polynomial $p\left( x \right)$ by $x - a$ and determine the quotient. After that, we factorize the quotient, using middle term splitting. The obtained factors are the other two factors of the given polynomial.

**Complete step by step answer:**

Given a polynomial is $f\left( x \right) = {x^3} + 2{x^2} - 51x + 108$ and a factor is $x + 9$. We have to determine the other two factors of the above given polynomial. For that, first we perform the long division of the given polynomial $f\left( x \right)$ by the factor $x + 9$, we get

$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{x^2} - 7x + 12 $

$ x + 9)\overline {{x^3} + 2{x^2} - 51x + 108} $

$ \,\,\,\,\,\,\,\,\,\,\,\,\underline {{x^3} + 9{x^2}} $

$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 7{x^2} - 51x $

$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, { - 16{x^2} - 63x} $

$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\overline{12x + 108 }$

$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline {12x + 108} $

$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 $

The quotient after dividing the given polynomial $f\left( x \right) = {x^3} + 2{x^2} - 51x + 108$ by $x + 9$ , we get ${x^2} - 7x + 12$ . Now we factorize this quadratic equation using middle term splitting.

First, we split the middle term $ - 7x$ as $ - 4x - 3x$ because the addition of both is equal to the middle term and the product is equal to the 1product of the first and last term of the quadratic equation. So, after middle term splitting, we get

$ \Rightarrow {x^2} - 4x - 3x + 12$

Now we take common from first two terms and last two terms, we get

$ \Rightarrow x\left( {x - 4} \right) - 3\left( {x - 4} \right)$

Now take $\left( {x - 4} \right)$ common, we get

$ \Rightarrow \left( {x - 4} \right)\left( {x - 3} \right)$

So the other two factors of the given polynomials are $\left( {x - 4} \right)$ and $\left( {x - 3} \right)$ .

**Note:**In middle term splitting, split the middle term such that the addition of the split terms is equal to the middle term and the product of the split terms is equal to the product of the first and last term of the quadratic equation.

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