Solution

(i) If $(x+1)$ is a factor of $p(x)=x^{3}+x^{2}+x+1$, then $p(-1)$ must be zero, otherwise $(x+1)$ is not a factor of $p(x)$.

$p(x)=x^{3}+x^{2}+x+1 p(-1)=$

$(-1)^{3}+(-1)^{2}+(-1)+1$

$=-1+1-1-1=0$

Hence, $x+1$ is a factor of this polynomial.

(ii) If $(x+1)$ is a factor of $p(x)=x^{4}+x^{3}+x^{2}+x+1$, then $p(-1)$ must be zero, otherwise $(x+1)$ is not a factor of $p(x) \cdot p(x)=x^{4}+x^{3}+x^{2}+x+1 p(-1)=$

$(-1)^{4}+(-1)^{3}+(-1)^{2}+(-1)+1$

$=1-1+1-1+1=1$

As $p \neq 0,(-1)$

Therefore, $x+1$ is not a factor of this polynomial.

(iii) If $(x+1)$ is a factor of polynomial $p(x)=x^{4}+3 x^{3}+3 x^{2}+x+1$, then $p(-1)$ must be 0 , otherwise $(x+1)$ is not a factor of this polynomial. $p(-1)=(-1)^{4}+3(-1)^{3}+3(-1)^{2}+(-1)+1$

$=1-3+3-1+1=1$

As $p \neq 0,(-1)$

Therefore, $x+1$ is not a factor of this polynomial.

(iv) If $(x+1)$ is a factor of polynomial $p(x) x^{x^{3}-x^{2}-(2+\sqrt{2}) x+\sqrt{2}}$, then $p(-1)$ must be 0 , otherwise $(x+1)$ is not a factor of this polynomial.

$ \begin{aligned} p(-1) & =(-1)^{3}-(-1)^{2}-(2+\sqrt{2})(-1)+\sqrt{2} \\ & =-1-1+2+\sqrt{2}+\sqrt{2} \\ & =2 \sqrt{2} \end{aligned} $

As $p \neq 0,(-1)$

Therefore, $(x+1)$ is not a factor of this polynomial.

Solution

(i) If $g(x)=x+1$ is a factor of the given polynomial $p(x)$, then $p(-1)$ must be zero. $p(x)=2 x^{3}+x^{2}-2 x-1 p(-1)=2(-1)^{3}+(-1)^{2}-2(-1)-1$

$=2(-1)+1+2-1=0$

Hence, $g(x)=x+1$ is a factor of the given polynomial.

(ii) If $g(x)=x+2$ is a factor of the given polynomial $p(x)$, then $p(-2)$ must be 0.

$ \begin{aligned} & p(x)=x^{3}+3 x^{2}+3 x+1 p(-2)= \\ & (-2)^{3}+3(-2)^{2}+3(-2)+1 \\ & =-8+12-6+1 \\ & =-1 \end{aligned} $

As $p \neq 0,(-2)$

Hence, $g(x)=x+2$ is not a factor of the given polynomial.

(iii) If $g(x)=x-3$ is a factor of the given polynomial $p(x)$, then $p(3)$ must be 0.

$ \begin{aligned} & p(x)=x^{3}-4 x^{2}+x+6 p(3) \\ & =(3)^{3}-4(3)^{2}+3+6=27 \\ & -36+9=0 \end{aligned} $

Hence, $g(x)=x-3$ is a factor of the given polynomial.

Solution

If $x-1$ is a factor of polynomial $p(x)$, then $p(1)$ must be 0 .

$ \begin{aligned} & \text{ (i) } p(x)=x^{2}+x+kp(1) \\ & =0 \\ & \Rightarrow(1) \quad 2+1+k=0 \\ & \Rightarrow(2)+k=0 \Rightarrow k \\ & =-2 \end{aligned} $

Therefore, the value of $k$ is -2 .

(ii)

$ p(x)=2 x^{2}+k x+\sqrt{2} $

$p(1)=0$

$\Rightarrow 2(1)^{2}+k(1)+\sqrt{2}=0$

$\Rightarrow 2+k+\sqrt{2}=0$

$\Rightarrow k=-2-\sqrt{2}=-(2+\sqrt{2})$

Therefore, the value of $k$ is $-(2+\sqrt{2})$.

(iii) $p(x)=k x^{2}-\sqrt{2} x+1$

$p(1)=0$

$\Rightarrow k(1)^{2}-\sqrt{2}(1)+1=0$

$\Rightarrow k-\sqrt{2}+1=0$

$\Rightarrow k=\sqrt{2}-1$

Therefore, the value of $k$ is $\sqrt{2}-1$.

(iv) $p(x)=k x^{2}-3 x+k$

$\Rightarrow p(1)=0 \Rightarrow k(1)^{2}-$

$3(1)+k=0 \Rightarrow k-3$

$+k=0$

$\Rightarrow \quad 2 k-3=0$

$\Rightarrow k=\frac{3}{2}$

Therefore, the value of $k$ is $\frac{3}{2}$.

Solution

(i) $12 x^{2}-7 x+1$

We can find two numbers such that $pq=12 \times 1=12$ and $p+q=-7$. They are $p=-4$ and $q=-3$.

Here, $12 x^{2}-7 x+1=12 x^{2}-4 x-3 x+1$

$=4 x(3 x-1)-1(3 x-1)$ $=(3 x-1)(4 x-1)$

(ii) $2 x^{2}+7 x+3$

We can find two numbers such that $p q=2 \times 3=6$ and $p+q=7$.

They are $p=6$ and $q=1$.

Here, $2 x^{2}+7 x+3=2 x^{2}+6 x+x+3$

$=2 x(x+3)+1(x+3)=(x$

$+3)(2 x+1)$

(iii) $6 x^{2}+5 x-6$

We can find two numbers such that $p q=-36$ and $p+q=5$.

They are $p=9$ and $q=-4$.

Here,

$6 x^{2}+5 x-6=6 x^{2}+9 x-4 x-6$

$=3 x(2 x+3)-2(2 x+3)$

$=(2 x+3)(3 x-2)$

(iv) $3 x^{2}-x-4$

We can find two numbers such that $pq=3 \times(-4)=-12$ and $p+q=-1$.

They are $p=-4$ and $q=3$.

Here,

$ \begin{aligned} & 3 x^{2}-x-4=3 x^{2}-4 x+3 x-4 \\ & =x(3 x-4)+1(3 x-4)= \\ & (3 x-4)(x+1) \text{ Question 5: } \end{aligned} $

Solution

(i) Let $p(x)=x^{3}-2 x^{2}-x+2$

All the factors of 2 have to be considered. These are $\pm 1, \pm 2$.

By trial method, $p(2)=(2)^{3}-2(2)^{2}-2+2$

$=8-8-2+2=0$

Therefore, $(x-2)$ is factor of polynomial $p(x)$.

Let us find the quotient on dividing $x^{3}-2 x^{2}-x+2$ by $x-2$.

By long division,

$$ \begin{matrix} x^2-3 x+2 \\ x + 1 \overline{) { x ^ { 3 } - 2 x ^ { 2 } - x + 2 }} \\ x^{3}+x^{2} \\ \hline - \\ \hline 3 x^{2}-x+2 \\ -3 x^{2}-3 x \\ +\quad+ \\ 2 x+2 \\ 2 x+2 \\ \end{matrix} $$

It is known that,

Dividend $=$ Divisor $\times$ Quotient + Remainder

$\therefore x^{3}$

$-2 x^{2}-x+2=(x+1)(x^{2}-3 x+2)+0=$

$(x+1)[x^{2}-2 x-x+2]$

$=(x+1)[x(x-2)-1(x-2)]$

$=(x+1)(x-1)(x-2)$

$=(x-2)(x-1)(x+1)$

(ii) Let $p(x)=x^{3}-3 x^{2}-9 x-5$

All the factors of 5 have to be considered. These are $\pm 1, \pm 5$.

By trial method, $p(-1)=(-1)^{3}-3(-1)^{2}-9(-1)-5$

$=-1-3+9-5=0$

Therefore, $x+1$ is a factor of this polynomial.

Let us find the quotient on dividing $x^{3}+3 x^{2}-9 x-5$ by $x+1$.

By long division,

$ \begin{matrix} x^{2}-4 x-5 \\ x + 1 )\overline { x ^ { 3 } - 3 x ^ { 2 } - 9 x - 5 } \\ x^{3}+x^{2} \\ -\quad- \\ \hline 4 x^{2}-9 x-5 \\ -4 x^{2}-4 x \\ +\quad+ \\ \hline \begin{matrix} -5 x-5 \\ -5 x-5 \\ + \end{matrix} \\ \hline 0 \end{matrix} $

It is known that,

$ \begin{aligned} & \text{ Dividend }=\text{ Divisor } \times \text{ Quotient }+ \text{ Remainder } \therefore x^{3} \\ & -3 x^{2}-9 x-5=(x+1)(x^{2}-4 x-5)+0 \\ & =(x+1)(x^{2}-5 x+x-5) \\ & =(x+1)[(x(x-5)+1(x-5)] \\ & =(x+1)(x-5)(x+1) \\ & =(x-5)(x+1)(x+1) \end{aligned} $

(iii) Let $p(x)=x^{3}+13 x^{2}+32 x+20$

All the factors of 20 have to be considered. Some of them are $\pm 1$,

$\pm 2, \pm 4, \pm 5 \ldots \ldots$ By trial method, $p(-1)$

$=(-1)^{3}+13(-1)^{2}+$

$ \begin{aligned} & 32(-1)+20 \\ & =-1+13-32+20 \\ & =33-33=0 \end{aligned} $

As $p(-1)$ is zero, therefore, $x+1$ is a factor of this polynomial $p(x)$.