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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