If $\left(x^{51}+51\right)$ is divided by
[question] Question. If $\left(x^{51}+51\right)$ is divided by $(x+1)$ then the remainder is (a) 0 (b) 1 (c) 49 (d) 50 [/question] [solution] Solution: Let $f(x)=x^{51}+51$ By remainder theorem, when f(x) is divided by (x + 1), then the remainder = f(−1). Putting x = −1 in f(x), we get $f(-1)=(-1)^{51}+51=-1+51=50$ ∴ Remainder = 50 Thus, the remainder when $\left(x^{51}+51\right)$ is divided by $(x+1)$ is 50 Hence, the correct answer is option (d). [/solution]...
If $p(x)=5 x-4 x^{2}+3$ then $p(-1)=?$
[question] Question. If $p(x)=5 x-4 x^{2}+3$ then $p(-1)=?$ (a) 2 (b) –2 (c) 6 (d) –6 [/question] [solution] Solution: $p(x)=5 x-4 x^{2}+3$ Putting $x=-1$ in $p(x)$, we get $p(-1)=5 \times(-1)-4 \times(-1)^{2}+3=-5-4+3=-6$ Hence, the correct answer is option (d). [/solution]...
If both $(x-2)$ and $\left(x-\frac{1}{2}\right)$ are factors of $p x^{2}+5 x+r$,
[question] Question. If both $(x-2)$ and $\left(x-\frac{1}{2}\right)$ are factors of $p x^{2}+5 x+r$, prove that $p=r$. [/question] [solution] Solution: Let $f(x)=p x^{2}+5 x+r$ It is given that $(x-2)$ is a factor of $f(x)$. Using factor theorem, we have $f(2)=0$ $\Rightarrow p \times\left(\frac{1}{2}\right)^{2}+5 \times \frac{1}{2}+r=0$ $\Rightarrow \frac{p}{4}+r=-\frac{5}{2}$ $\Rightarrow p+4 r=-10 \quad \ldots \ldots(2)$ From (1) and (2), we have $4 p+r=p+4 r$ $\Rightarrow 4 p-p=4 r-r$ $\Rig...
If $\left(x^{3}+a x^{2}+b x+6\right)$ has $(x-2)$ as a factor and leaves a remainder 3 when divided by
[question] Question. If $\left(x^{3}+a x^{2}+b x+6\right)$ has $(x-2)$ as a factor and leaves a remainder 3 when divided by $(x-3)$, find the values of $a$ and $b$. [/question] [solution] Solution: Let: $f(x)=x^{3}+a x^{2}+b x+6$ $(x-2)$ is a factor of $f(x)=x^{3}+a x^{2}+b x+6$ $\Rightarrow f(2)=0$ $\Rightarrow 2^{3}+a \times 2^{2}+b \times 2+6=0$ $\Rightarrow 14+4 a+2 b=0$ $\Rightarrow 4 a+2 b=-14$ $\Rightarrow 2 a+b=-7 \quad \ldots(1)$ Now, $x-3=0 \Rightarrow x=3$ By the factor theorem, we ca...
Show that $(p-1)$ is a factor of $\left(p^{10}-1\right)$ and also of $\left(p^{11}-1\right)$.
[question] Question. Show that $(p-1)$ is a factor of $\left(p^{10}-1\right)$ and also of $\left(p^{11}-1\right)$. [/question] [solution] Solution: Let $f(p)=p^{10}-1$ and $g(p)=p^{11}-1$ Putting $p=1$ in $f(p)$, we get $f(1)=1^{10}-1=1-1=0$ Therefore, by factor theorem, $(p-1)$ is a factor of $\left(p^{10}-1\right)$. Now, putting $p=1$ in $g(p)$, we get $g(1)=1^{11}-1=1-1=0$ Therefore, by factor theorem, $(p-1)$ is a factor of $\left(p^{11}-1\right)$. [/solution]...