# How to apply the remainder Theorem?

## () Can someone please explain to me how to do 3 c and 3.7? Thank you so much!

Mar 26, 2017

#### Explanation:

According to remainder theorem if $f \left(x\right)$ is divided by $\left(x - a\right)$, then remainder is $f \left(a\right)$. Therefore if $\left(x - a\right)$ is a factor of $f \left(x\right)$, $f \left(a\right) = 0$.

Now coming to questions raised by you solution is given seriatim.

(a) As $f \left(x\right) = {x}^{n} - {a}^{n}$, dividing by $x - a$ gives a remainder $f \left(a\right) = {a}^{n} - {a}^{n} = 0$. Hence ${x}^{n} - {a}^{n}$ is divisible by $x - a$.

(b.I) As $f \left(x\right) = {x}^{n} + {a}^{n}$, dividing by $x + a$ gives a remainder $f \left(a\right) = {\left(- a\right)}^{n} + {a}^{n} = {\left(- 1\right)}^{n} {a}^{n} + {a}^{n}$. This will be $0$ only if $n$ is odd. Hence the condition for $x + a$ to be a factor of ${x}^{n} - {a}^{n}$ is $n$ is odd.

(b.II) As $f \left(x\right) = {x}^{n} - {a}^{n}$, dividing by $x + a$ gives a remainder $f \left(a\right) = {\left(- a\right)}^{n} - {a}^{n} = {\left(- 1\right)}^{n} {a}^{n} - {a}^{n}$. This will be $0$ only if $n$ is even. Hence condition for $x + a$ to be a factor of ${x}^{n} - {a}^{n}$ is $n$ is even.

3 c As $f \left(x\right) = 2 {x}^{3} + {x}^{2} - 5 x + 2$, the factors are of the type $x - a$, where $a$ is a factor of $\frac{2}{2}$ ($\frac{p}{q}$ where $p$ is constant term and $q$ is coefficient of highest power of $x$) i.e. $\pm 1$ or $\pm 2$ or $\pm \frac{1}{2}$. As $f \left(1\right) = 0$, $f \left(2\right) = 0$ and $f \left(\frac{1}{2}\right) = 0$, hence factors are $x - 1$, $x - 2$ and $2 x - 1$ i.e. $2 {x}^{3} + {x}^{2} - 5 x + 2 = \left(x - 1\right) \left(x - 2\right) \left(2 x - 1\right)$.

3.7 When $P \left(x\right)$ is divided by $\left(x - 1\right)$, remainder is $2$, hence $P \left(1\right) = 2$ and as when $P \left(x\right)$ is divided by $\left(x - 2\right)$, remainder is $3$, hence $P \left(2\right) = 3$.

(a) As $P \left(x\right) = \left(x - 1\right) \left(x - 2\right) Q \left(x\right) + a x + b$,

$P \left(1\right) = 2$ gives us $a + b = 2$ and $P \left(2\right) = 3$ gives us $2 a + b = 3$.

Subtracting former from latter, we get $a = 1$ and hence $b = 1$

(b.I) As $P \left(x\right) = \left(x - 1\right) \left(x - 2\right) Q \left(x\right) + a x + b$ and $P \left(x\right)$ is a cubic polynomial, with coefficient of ${x}^{3}$ as $1$, it is of the type

$P \left(x\right) = \left(x - 1\right) \left(x - 2\right) \left(x - k\right) + x + 1$. Now as $- 1$ is a solution to $P \left(x\right) = 0$, we have

$P \left(- 1\right) = \left(- 1 - 1\right) \left(- 1 - 2\right) \left(- 1 - k\right) - 1 + 1 = - 6 - 6 k = 0$

i.e. $k = - 1$ and $P \left(x\right) = \left(x - 1\right) \left(x - 2\right) \left(x + 1\right) + x + 1$ or

$P \left(x\right) = {x}^{3} - 2 {x}^{2} - 4 x - 1$

(b.II) It is apparent that as $P \left(x\right) = \left(x - 1\right) \left(x - 2\right) \left(x + 1\right) + x + 1$, $\left(x + 1\right)$ is a factor of $P \left(x\right)$ and

$P \left(x\right) = \left(x - 1\right) \left(x - 2\right) \left(x + 1\right) + x + 1$

= $\left(x + 1\right) \left({x}^{2} - 3 x + 2 + 1\right) = \left(x + 1\right) \left({x}^{2} - 3 x + 3\right)$

As discriminant in ${x}^{2} - 3 x + 3$ is ${3}^{2} - 4 \times 1 \times 3 = 9 - 12 = - 3$,

there is no other real factor.