Question

Factorize the given expression using the factor theorem. The expression is:
$$x^3-23x^2+142x-120$$

Answer 1

Hint: First, find all the factors of the constant term i.e. -120. Then for factor a, if p(a) =0 , then (x-a) is the factor of the polynomial p(x).

Complete step-by-step answer:
In the question, we have to factorize the polynomial $$p\left(x\right)=x^3-23x^2+142x-120$$.
So for that we have to find the factors of this polynomial $$p\left(x\right)$$ using the factor theorem.
Now, $$x-a$$ is the factor of the polynomial $$p\left(x\right)$$ if $$p\left(a\right)=0$$, and this is called the factor theorem.
Now, we will find the factors of -120. So the factor of -120 are:
$$\pm 1,\pm 2,\pm 3,\pm 4,\pm 5,\pm 6,\pm 8,\pm 10,\pm 12,\pm 15,\pm 20,\pm 24,\pm 30,\pm 40,\pm 60,\pm 120$$
Next, we will first take 1, and check if $$p\left(1\right)$$ is zero or not, if we get $$p\left(1\right)=0$$, then $$x-1$$will be the factor of this polynomial $$p\left(x\right)=x^3-23x^2+142x-120$$
Now, the value of $$p\left(1\right)$$ is found as follows:
$$\begin{array}{rl}& \Rightarrow p\left(1\right)={\left(1\right)}^3-23{\left(1\right)}^2+142\left(1\right)-120\\ & \Rightarrow p\left(1\right)=0\end{array}$$
So we get $$p\left(1\right)=0$$ which means that $$x-1$$ is a factor of the polynomial $$x^3-23x^2+142x-120$$.
Next, we can write the polynomial $$p\left(x\right)=x^3-23x^2+142x-120$$, as $$p(x)=(x-1){\displaystyle \frac{x^3-23x^2+142x-120}{x-1}}$$
Next we will find the expression of $${\displaystyle \frac{x^3-23x^2+142x-120}{x-1}}$$, and this is done as follows:
$$\begin{array}{rl}& \Rightarrow {\displaystyle \frac{x^3-23x^2+142x-120}{x-1}}\\ & \Rightarrow {\displaystyle \frac{x^3-x^2-22x^2+22x+120x-120}{x-1}}\\ & \Rightarrow {\displaystyle \frac{x^2(x-1)-22x(x-1)+120(x-1)}{x-1}}\\ & \Rightarrow x^2-22x+120\end{array}$$
So we can write the polynomial as
$$\begin{array}{rl}& \Rightarrow p(x)=(x-1){\displaystyle \frac{x^3-23x^2+142x-120}{x-1}}\\ & \Rightarrow p(x)=(x-1)(x^2-22x+120)\end{array}$$
Next, we will split the middle term of the quadratic expression$$(x^2-22x+120)$$, so as to get the factors. This is done as follows:
$$\begin{array}{rl}& \Rightarrow (x^2-22x+120)\\ & \Rightarrow (x^2-10x)+(-12x+120)\\ & \Rightarrow x(x-10)-12(x-10)\\ & \Rightarrow (x-10)(x-12)\end{array}$$
So finally the polynomial is written in the factor form as follows:
$$\begin{array}{rl}& \Rightarrow p\left(x\right)=x^3-23x^2+142x-120\\ & \Rightarrow p(x)=(x-1)(x^2-22x+120)\\ & \Rightarrow p(x)=(x-1)(x-10)(x-12)\end{array}$$
So the required factor form is $$p(x)=(x-1)(x-10)(x-12)$$.

Note: Make sure there is no calculation error while finding the value of $$p\left(1\right)$$, here we have to replace $$x=1$$ in the polynomial $$p\left(x\right)$$.