# Factor Theorem with Proof and Examples

Factor Theorem : If $$p(x)$$ is a polynomial of degree $$n \geq 1$$ and $$a$$ is any real number, then $$x-a$$ is a factor of $$p(x)$$, if  and only if $$p(a)=0$$.

Proof: By the Remainder Theorem, $$p(x)=(x-a) q(x)+p(a)$$.
(i) If $$p(a)=0$$, then $$p(x)=(x-a) q(x)$$, which shows that $$x-a$$ is a factor of $$p(x)$$.
(ii) Since $$x-a$$ is a factor of $$p(x), p(x)=(x-a) g(x)$$ for same polynomial $$g(x)$$. In this case, $$p(a)=(a-a) g(a)=0$$.

Solution : The zero of $$x+2$$ is $$-2$$. Let $$p(x)=x^3+3 x^2+5 x+6$$ and $$s(x)=2 x+4$$ Then, $$\quad p(-2)=(-2)^3+3(-2)^2+5(-2)+6$$

\begin{aligned} &=-8+12-10+6 \\ &=0 \end{aligned}
So, by the Factor Theorem, $$x+2$$ is a factor of $$x^3+3 x^2+5 x+6$$.
Again,
$s(-2)=2(-2)+4=0$
So, $$x+2$$ is a factor of $$2 x+4$$. In fact, you can check this without applying the Factor Theorem, since $$2 x+4=2(x+2)$$.

Solution : As $$x-1$$ is a factor of $$p(x)=4 x^3+3 x^2-4 x+k, p(1)=0$$
Now,
$p(1)=4(1)^3+3(1)^2-4(1)+k$
So,
\begin{aligned} 4+3-4+k &=0 \\ k &=-3 \end{aligned}
We will now use the Factor Theorem to factorise some polynomials of degree 2 and 3 . You are already familiar with the factorisation of a quadratic polynomial like $$x^2+l x+m$$. You had factorised it by splitting the middle term $$l x$$ as $$a x+b x$$ so that $$a b=m$$. Then $$x^2+l x+m=(x+a)(x+b)$$. We shall now try to factorise quadratic polynomials of the type $$a x^2+b x+c$$, where $$a \neq 0$$ and $$a, b, c$$ are constants.
Factorisation of the polynomial $$a x^2+b x+c$$ by splitting the middle term is as follows:

Let its factors be $$(p x+q)$$ and $$(r x+s)$$. Then
$a x^2+b x+c=(p x+q)(r x+s)=p r x^2+(p s+q r) x+q s$
Comparing the coefficients of $$x^2$$, we get $$a=p r$$.
Similarly, comparing the coefficients of $$x$$, we get $$b=p s+q r$$.
And, on comparing the constant terms, we get $$c=q s$$.
This shows us that $$b$$ is the sum of two numbers $$p s$$ and $$q r$$, whose product is $$(p s)(q r)=(p r)(q s)=a c$$.

Therefore, to factorise $$a x^2+b x+c$$, we have to write $$b$$ as the sum of two numbers whose product is $$a c$$.

Solution 1 : (By splitting method) : If we can find two numbers $$p$$ and $$q$$ such that $$p+q=17$$ and $$p q=6 \times 5=30$$, then we can get the factors.

So, let us look for the pairs of factors of 30 . Some are 1 and 30,2 and 15,3 and 10,5 and 6. Of these pairs, 2 and 15 will give us $$p+q=17$$.
So, $$6 x^2+17 x+5=6 x^2+(2+15) x+5$$
\begin{aligned} &=6 x^2+2 x+15 x+5 \\ &=2 x(3 x+1)+5(3 x+1) \\ &=(3 x+1)(2 x+5) \end{aligned}
Solution 2 : (Using the Factor Theorem)
$$6 x^2+17 x+5=6\left(x^2+\frac{17}{6} x+\frac{5}{6}\right)=6 p(x)$$, say. If $$a$$ and $$b$$ are the zeroes of $$p(x)$$, then $$6 x^2+17 x+5=6(x-a)(x-b)$$. So, $$a b=\frac{5}{6}$$. Let us look at some possibilities for $$a$$ and
b. They could be $$\pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{5}{3}, \pm \frac{5}{2}, \pm 1$$. Now, $$p\left(\frac{1}{2}\right)=\frac{1}{4}+\frac{17}{6}\left(\frac{1}{2}\right)+\frac{5}{6} \neq 0$$. But $$p\left(\frac{-1}{3}\right)=0$$. So, $$\left(x+\frac{1}{3}\right)$$ is a factor of $$p(x)$$. Similarly, by trial, you can find that $$\left(x+\frac{5}{2}\right)$$ is a factor of $$p(x)$$
Therefore, $$\quad 6 x^2+17 x+5=6\left(x+\frac{1}{3}\right)\left(x+\frac{5}{2}\right)$$
$$=6\left(\frac{3 x+1}{3}\right)\left(\frac{2 x+5}{2}\right)$$
$$=(3 x+1)(2 x+5)$$

Solution : Let $$p(y)=y^2-5 y+6$$. Now, if $$p(y)=(y-a)(y-b)$$, you know that the constant term will be $$a b$$. So, $$a b=6$$. So, to look for the factors of $$p(y)$$, we look at the factors of 6 .
The factors of 6 are 1,2 and 3 .
Now, $$\quad p(2)=2^2-(5 \times 2)+6=0$$

So, $$y-2$$ is a factor of $$p(y)$$.
Also, $$p(3)=3^2-(5 \times 3)+6=0$$
So, $$y-3$$ is also a factor of $$y^2-5 y+6$$.
Therefore, $$y^2-5 y+6=(y-2)(y-3)$$
Note that $$y^2-5 y+6$$ can also be factorised by splitting the middle term $$-5 y$$.

Solution: Let $$p(x)=x^3-23 x^2+142 x-120$$
We shall now look for all the factors of $$-120$$. Some of these are $$\pm 1, \pm 2, \pm 3$$, $$\pm 4, \pm 5, \pm 6, \pm 8, \pm 10, \pm 12, \pm 15, \pm 20, \pm 24, \pm 30, \pm 60$$.
By trial, we find that $$p(1)=0$$. So $$x-1$$ is a factor of $$p(x)$$.
Now we see that $$x^3-23 x^2+142 x-120=x^3-x^2-22 x^2+22 x+120 x-120$$ $$=x^2(x-1)-22 x(x-1)+120(x-1) \quad$$

$$=(x-1)\left(x^2-22 x+120\right) \quad$$ [Taking $$(x-1)$$ common]
We could have also got this by dividing $$p(x)$$ by $$x-1$$.
Now $$x^2-22 x+120$$ can be factorised either by splitting the middle term or by using the Factor theorem. By splitting the middle term, we have:
\begin{aligned} x^2-22 x+120 &=x^2-12 x-10 x+120 \\ &=x(x-12)-10(x-12) \\ &=(x-12)(x-10) \end{aligned}
So, $$\quad x^3-23 x^2-142 x-120=(x-1)(x-10)(x-12)$$

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