Question

Using Factor theorem, factorise: $p\left(x\right)=x^3-23x^2+142x-120$.

Answer 1

Factor Theorem: -$p\left(x\right)$ be a polynomial of degree greater than or equal to $1$ and $a$ be a real number. $p\left(a\right)=0$, then $(x-a)$ is a factor of $p\left(x\right)$.

Finding the factors of equation $x^3-23x^2+142x-120$.

Step 1: Given, the equation, $p\left(x\right)=x^3-23x^2+142x-120$

Step 2: Calculating 1st factor by Factor's Therorem

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:

$$\begin{gathered} p\left(1\right)=1³-23\times 1²+142\times 1-120 \\ \Rightarrow p\left[1\right]=1-23+142-120 \\ \Rightarrow p\left[1\right]=143-143 \\ \Rightarrow p\left[1\right]=0 \end{gathered}$$

So$(x-1)$ is a factor.

Step 3: Caclculating other factors from middle split theorem

Now, to divide $(x-1)$ with $x^3-23x^2+142x-120$ by doing this we will get$x²-22x+120$

Find the other factors of $x^3-23x^2+142x-120$ we can split the middle term

$$\begin{gathered} x^3-23x^2+142x-120 \\ =x²-12x+(-10x)+120 \\ =x(x-12)-10(x-12) \\ =(x-10)(x-12) \end{gathered}$$

Hence, the factors of $x^3-23x^2+142x-120$ are $(x-1)(x-10)(x-12)$