Question 511 Mark
If $x^{140} + 2x^{151} + k$ divisible by $x + 1,$ then the value of $k$ is:
AnswerZero of $(x + 1) = -1.$
By Remainder theorem,
$P(x) = x^{140} + 2x^{151} + k$
$p(1) = (-1)^{140} + 2(-1)^{151} + k = 0$
$1 + 2(-1) + k = 0$
$1 - 2 + k = 0$
$-1 + k = 0$
$k = 1.$
View full question & answer→Question 521 Mark
When $p(x) = 4x^3 - 12x^2 + 11x -5$ is divided by $(2x - 1),$ the remainder is:
Answer$2\text{x}-1=0$
$\Rightarrow\text{x}=\frac{1}{2}$
By the remainder theorem, we know that when $p(x)$ is divided by $(2x - 1),$ the remainder is $\text{p}\Big(\frac{1}{2}\Big)$
Now, we have:
$\text{p}\Big(\frac{1}{2}\Big)=4\times\Big(\frac{1}{2}\Big)^3-12\times\Big(\frac{1}{2}\Big)^2+11\times\frac{1}{2}-5$
$=\frac{1}{2}-3+\frac{11}{2}-5$
$=-2$
View full question & answer→Question 531 Mark
If $\text{x}^2+\frac{1}{\text{x}^2}=102,$ than $\text{x}-\frac{1}{\text{x}}=$
Answer- 10
Solution:
$\text{x}^2+\frac{1}{\text{x}^2}=102,$
$\Rightarrow(\text{x}^2)+\Big(\frac{1}{\text{x}^2}\Big)-2\times\text{x}\times\frac{1}{\text{x}}=102-2\times\text{x}\times\frac{1}{\text{x}}$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=100$
$\Rightarrow\text{x}-\frac{1}{\text{x}}=\sqrt{100}=10$
View full question & answer→Question 541 Mark
Write the correct answer in the following: $If\ x^{51} + 51$ is divided by $x + 1,$ the remainder is.
AnswerIf $p(x)$ is divided by $x + a,$ then the remainder is $p(-a).$
Here $p(x) = x^{51} + 51$ is divided by $x + 1,$ then
$x = -1$
Remainder $= p(-1) = (-1)^{51} + 51 = 50 = -1 + 51 = 50$
View full question & answer→Question 551 Mark
Write the correct answer in the following: If $x + 1$ is a factor of the polynomial $2x^2 + kx,$ then the value of $k$ is.
AnswerLet $p(x) = 2x^2 + kx$
Since$, (x + 1)$ is a factor of $p(x),$ then
$p(-1) = 0$
$2(-1)^2 + k(-1) = 0$
$\Rightarrow 2 - k = 0$
$\Rightarrow k = 2$
View full question & answer→Question 561 Mark
The Zero of the polynomial $(x - 2)^2 - (x + 2)^2$ is:
Answer$(x - 2)^2 - (x + 2)^2$
$= (x - 2 + x + 2) (x - 2 - x - 2) [$Using identity $a^2 - b^2 = (a + b) (a - b)]$
$= (2x) (-4)$
$= -8x$
Then the zero is,
$-8 = 0$
$\Rightarrow x = 0$
View full question & answer→Question 571 Mark
If $p(x) = 5x - 4x^2 + 3$ then $p(-1) =$ ?
Answer$p(x) = 5x - 4x^2 + 3$
Putting $x = -1$ in$ p(x),$ we get
$p(-1) = 5 \times (-1)-4 \times (-1)^2 + 3 = -5 - 4 + 3 = -6$
View full question & answer→Question 581 Mark
The value of $(249)^2 - (248)^2$ is:
Answer$(249)^2 - (248)^2$
We know
$a^2 - b^2 = (a + b)(a - b)$
So$,$
$(249)^2 - (248)^2$
$(249 - 248)(249 + 248)$
$= 497$
View full question & answer→Question 591 Mark
Write the correct answer in the following:
If p(x) = x + 3, then p(x) + p(-x) is equal to.
Answer- 6
Solution:
We have p(x) = x + 3, then
p(-x) = -x + 3
Therefore, p(x) + p(-x) = x + 3 + (-x + 3) = x + 3 - x + 3 = 6
View full question & answer→Question 601 Mark
The product $(x^2 - 1)(x^4 + x^2 + 1)$ is equal to:
AnswerGiven expression is$ (x^2 - 1)(x^4 + x^2 + 1)$
Let $x^2 = A$ and $1 = B$
Then, we have
$(A - B)(A^2 + AB + B^2)$
$= A^3 - B^3$
$= (X^2)^3 - (1)^3$
$= X^6 - 1$
View full question & answer→Question 611 Mark
Which of the following is a quadratic polynomial?
Answer$x^2 + 5x + 4$ is a polynomial of deree $2.$
So, it is a quadratic polynomial.
View full question & answer→Question 621 Mark
Write the correct answer in the following:
Zero of the polynomial p(x) = 2x + 5 is.
Answer- $-\frac{5}{2}$
Solution:
Finding a zero of p(x) is the same as solving an equation p(x) = 0.
Now, p(x) = 0 ⇒ 2x + 5 = 0,
2x = -5
Which give us $\text{x}=-\frac{5}{2}.$
Therefore, $-\frac{5}{2}$ is the zero of the polynomial.
View full question & answer→Question 631 Mark
Write the correct answer in the following: If $a + b + c = 0,$ then $a^3 + b^3 + c^3$ is equal to.
AnswerNow, $a^3 + b^3 + c^3 = (a + b + c)(a^2 + b^2 + c^2 - ab - be - ca) + 3abc$
$[$Using identit $y, a^3 + b^3 + c^3 – 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - be - ca)] = 0 + 3abc$
$\therefore a + b + c = 0,$ given
$a^3 + b^3 + c^3 = 3abc$
View full question & answer→Question 641 Mark
A polynomial of degree 3 in x has at most.
Answer- 3 terms
Solution:
3 terms of not more than the power of 3
View full question & answer→Question 651 Mark
The zeros of the polynomial $p(x) = 2x^2 + 5x - 3$ are.
AnswerThe given polynomial is $p(x) = 2x^2 + 5x - 4$
Putting $\text{x}=\frac{1}{2}$ in $p(x),$ we get
$\text{p}\Big(\frac{1}{2}\Big)=2\times\Big(\frac{1}{2}\Big)^2+5\times\frac{1}{2}-3$
$=\frac{1}{2}+\frac{5}{2}-3=3-3=0$
Putting $x = -3$ in$ p(x),$ we get
$\text{p}(-3)=2\times(-3)^2+5\times(-3)-3$
$= 18-15-3$
$= 0$
Therefore$, x = -3$ is a zero of the polynomial $p(x)$
Thus, $\frac{1}{2}$ and $-3$ are the zeros of the given polynomial $p(x).$
View full question & answer→Question 661 Mark
When $p(x) = x^4 + 2x^3 - 3x^2 + x - 1$ is divided by $(x - 2),$ the remainder is:
Answer$x^4 + 2x^3 - 3x^2 + x - 1$
Using remainder theorem,
$= (2)^4 + 2(2)^3 - 3(2)^2 + 2 - 1$
$= 16 + 16 - 12 + 2 - 1$
$= 34 - 13$
$= 21$
View full question & answer→Question 671 Mark
The value of $\frac{(2.3)^3-0.027}{(2.3)^2+0.69+0.09},$ is:
Answer- 2
Solution:
$\frac{(2.3)^3-0.027}{(2.3)^2+0.69+0.09}$
$=\frac{(2.3)^3-(0.3)^3}{(2.3)^2+(0.3)^3+(2.3)(0.3)}$
$=\frac{(2.3 - 0.3)\{(2.3)^2+(0.3)^2+(2.3)(0.3)\}}{((2.3)^2+(0.3)^2+(2.3)(0.3))}$
$=2.3 - 0.3$
$=2$
Hence, correct option is (a).
View full question & answer→Question 681 Mark
If $x + 2$ and $x - 1$ are the factor of $x^3 + 10x^2 + mx + n,$ then the values of $m$ and $n$ are respectively.
AnswerIt is given $(x + 2)$ and $(x - 1)$ are the factors of the polynomial $f(x)= x^3 + 10x^2 + mx + n$
$i.e., f(-2) = 0$ and $f(1) = 0$
New,
f$(-2) = (-2)^3 + 10(-2)^2 + m(-2) + n = 0$
$-8 + 40 - 2m + n =0$
$\Rightarrow -2m + n = -32$
$\Rightarrow 2m - n = 32 ....(i)$
$f(1) = (1)^3 + 10(1)^2 + m(1) + n = 0$
$1 + 10 + m + n = 0$
$m + n = -11 .....(ii)$
Solving equation $(i)$ and $(ii)$ we get,
$m = 7$ and $n = -18$
View full question & answer→Question 691 Mark
If$ (x + y)^3 - (x - y)^3 - 6y(x^2 - y^2) = ky^2,$ then $k =$
AnswerLet$ x + y = A$ and $x - y = B$
Now$, (A - B)^3 = A^3 - B^3 - 3AB(A - B)$
$\Rightarrow [(x + y) - (x - y)]^3 = (x + y)^3 - (x - y)^3 - 3(x + y)(x - y)[(x + y) - (x - y)]$
$= (x + y)^3 - (x - y)^3 - 3(x^2 - y^2)(2y)$
$= (x + y)^3 - (x - y)^3 - 6y(x^2 - y^2)$
But$, (x + y)^3 - (x - y)^3 - 6y(x^2 - y^2) = ky^3$
$\Rightarrow [(x + y) - (x - y)]^3 = (2y)^3 = k8y^3$
$\Rightarrow (2y)^3 = ky^3$
$\Rightarrow 8y^3 = ky^3$
$\Rightarrow k = 8$
View full question & answer→Question 701 Mark
If $p(x) = 5x - 4x^2 + 3$ then $p(-1) =$ ?
Answer$P(x) = 5x - 4x^2 + 3$
$\Rightarrow p(-1) = 5(-1) - 4(-1)^2 + 3$
$= -5 - 4 + 3$
$= -6$
View full question & answer→Question 711 Mark
Write the correct answer in the following:
Which one of the following is a polynomial?
Answer- $\text{x}^2+\frac{3\text{x}^{\frac{3}{2}}}{\sqrt{\text{x}}}$
Solution:
- Now, $\frac{\text{x}^2}{2}-\frac{2}{\text{x}^2}=\frac{\text{x}^2}{2}-2\text{x}^{-2},$ it is not a polynomial, because exponent of x is -2 which is not a whole number.
- Now, $\sqrt{2\text{x}-1}=\sqrt{\text{2}\text{x}}^{\frac{1}{2}}-1, $ it is not a polynomial, because exponent of x is $-\frac{1}{2}$ which is not a whole number.
- Now, $\text{x}^2+\frac{3\text{x}^{\frac{3}{2}}}{\sqrt{\text{x}}}=\text{x}^2+3\text{x}^{\frac{3}{2}-\frac{1}{2}}=\text{x}^2+3\text{x}^{\frac{2}{2}}=\text{x}^2+3\text{x},$ it is not a polynomial, because exponent of x is which is a whole number.
- $\frac{\text{x}-1}{\text{x}+1},$ it is not a polynomial because it is a rational function.
View full question & answer→Question 721 Mark
The zeroes of the polynomial p(x) = x(x - 2) (x + 3) are:
Answer- 0, 2, -3
Solution:
p(x) = x(x - 2) (x + 3)
⇒ x = 0 and x - 2 = 0 and x + 3 = 0
⇒ x = 0, x = 2 and x = -3
Therefore, the zeroes are 0, 2, -3
View full question & answer→Question 731 Mark
The factors of $a^2 - 1 - 2x - x^2,$ are.
AnswerThe given expression to be factorized is $a^2 - 1 - 2x - x^2$
Take common $-1$ from the last three terms and then we have
$a^2 - 1 - 2x - x^2 = a^2 - (1 + 2x + x^3)$
$= a^2 - {(1)^2 + 2.1 \times x + (x)^2}$
$= a^2 - (1 + x)^2$
$= (a)^2- (1 + x)^2$
$= {a + (1 + x)} {a - (1 + x)}$
$= (a + 1 + x) (a - 1 - x)$
$= (a + x + 1) (a - x - 1)$
View full question & answer→Question 741 Mark
If both x - a and $\text{x}-\frac{1}{2}$ are the factors of $\text{px}^2 + 5\text{x} +\text{r},$ than:
Answer- p = r
Solution:
If both x - a and $\text{x}-\frac{1}{2}$ are the factors of $\text{px}^2 + 5\text{x} +\text{r},$ than f(2) = 0
$\Rightarrow\text{p}(2)^2+5(2)+\text{r}=0$
$\Rightarrow4\text{p}+10+\text{r} = 0$
$\Rightarrow4\text{p}+\text{r}=-10\ ...(\text{i})$
Also, $\text{f}\Big(\frac{1}{2}\Big)=0$
$\Rightarrow\text{p}\Big(\frac{1}{2}\Big)^2+5\Big(\frac{1}{2}\Big)+\text{r}=0$
$\Rightarrow\frac{\text{p}}{4}+\frac{5}{2}+\text{r}=0$
$\Rightarrow\text{p}+10+4\text{r}=0$
$\Rightarrow\text{p}+4\text{r}=-10\ ...{\text{ii}}$
From eq. (i) and eq. (ii), we get
$4\text{p}+\text{r}=\text{p}+4\text{r}$
$\Rightarrow3\text{p}=3\text{r}$
$\Rightarrow\text{p}=\text{r}$
View full question & answer→Question 751 Mark
If $49\text{a}^2-{\text{b}}=\Big(7\text{a}+\frac{1}{2}\Big)\Big(7\text{a}-\frac{1}{2}\Big),$ then the value of $b$ is:
Answer$\Big(7\text{a}+\frac{1}{2}\Big)\Big(7\text{a}-\frac{1}{2}\Big)=(7\text{a})^2-\Big(\frac{1}{2}\Big)^2$
$[$by using identity $(a + b)(a - b) = a^2 - b^2]$
$\Rightarrow\Big(7\text{a}+\frac{1}{2}\Big)\Big(7\text{a}-\frac{1}{2}\Big)=49\text{a}^2-\frac{1}{4}$
$\Rightarrow49\text{a}^2-\text{b}=49\text{a}^2-\frac{1}{4}$
$\Rightarrow\text{b}=\frac{1}{4}$
View full question & answer→Question 761 Mark
Which of the following expression is a monomial.
Answer$4x^3$ because monomial means only one term in an expression.
View full question & answer→Question 771 Mark
Write the correct answer in the following: The value of $249^2 - 248^2$ is.
Answer$(249)^2 - (248)^2 $
$= (249 + 248)(249 - 248) [(a)^2 - (b)^2 $
$= (a + b)(a - b)]$
$= (497)(1) = 497$
View full question & answer→Question 781 Mark
The zeros of the polynomial $p(x) = 3x^2 - 1$ are:
Answer$p(x) = 3x^2 - 1$
Now$, p(x) = 0$
$\Rightarrow 3x^2- 1 = 0$
$\Rightarrow 3x^2 = 1$
$\Rightarrow\text{x}^2=\frac{1}{3}$
$\Rightarrow\text{x}=\frac{1}{\sqrt3}\ \text{and}\ -\frac{1}{\sqrt3}$
View full question & answer→Question 791 Mark
If p(x) = x + 4 then p(x) + p(-x) = ?
Answer- 8
Solution:
p(x) = x + 4
p(-x) = -x + 4
p(x) + p(-x) = (x + 4) + (-x + 4)
= x + 4 - x + 4
= 8
View full question & answer→Question 801 Mark
Write the correct answer in the following: One of the zeroes of the polynomial $2x^2 + 7x - 4$ is.
AnswerLet$ p(x) = 2x^2 + 7x - 4$
$= 2x^2 + 8x - x- 4\ [$by splitting middle term$]$
$= 2x(x + 4) -1(x + 4)$
$=(2x - 1)(x + 4)$
For zeroes of $p(x),$ put $p(x) = 0$
$\Rightarrow (2x - 1)(x + 4) = 0$
$\Rightarrow 2x - 1 = 0$ and $x + 4 = 0$
$\Rightarrow\text{x}=\frac{1}{2}$ and $\text{x}=-4$
Hence, one of the zeroes of the polynomial $p(x)$ is $\frac{1}{2}.$
View full question & answer→Question 811 Mark
The value of $(102)^3$ is:
Answer$(102)^3 $
$= (100 + 2)^3$
$= (100)^3 + (2)^3 + 3 \times 100 \times 2(100 + 2)$
$= 1000000 + 8 + 60000 + 1200$
$= 1061208$
View full question & answer→Question 821 Mark
Which of the following is a linear polynomial?
Answer- x + 1
Solution:
A polynomial of degree 1 is called a linear polynomial.
Options (a), and (c) have degree 2,
so ther are quadratic polynomials.
option (d) has a negative power, so it is not a polynomial.
The degree of x + 1 is 1, so it is a linear polynomial.
View full question & answer→Question 831 Mark
If $x - 2$ is $a$ factor of $x^2 + 3ax - 2a,$ then $a =$
AnswerAs $(x - 2)$ is a factor of $f(x) = x^2 + 3ax - 2a$
$i.e., f(2) = 0$
$(2)^2 + 3a(2) - 2a = 0$
$4 + 6a - 2a = 0$
$= -1$
View full question & answer→Question 841 Mark
If both $(x + 2)$ and $(2x + 1)$ are factors of $ax^2 + 2x + b,$ than the value of $a - b$ is:
Answer$\text{p}(\text{x})=\text{ax}^2+2\text{x}+\text{b}$
$\Rightarrow\text{a}(-2)^2 + 2(-2) + b = 0$
$\Rightarrow4\text{a} - 4+\text{b}=0\ .... (\text{i})$
Also,
$\text{p}\Big(\frac{-1}{2}\Big)=0$
$\Rightarrow\text{a}\Big(\frac{-1}{2}\Big)^2+2\Big(\frac{-1}{2}\Big)+\text{b}=0$
$\Rightarrow\frac{\text{a}}{4}-1+\text{b}=0$
$\Rightarrow\text{a}-4+4\text{b}=0\ ...(\text{ii})$
Subtracting eq. $(ii)$ from eq. $(i),$ we get
$3\text{a}+0-3\text{b}=0$
$\Rightarrow3(\text{a}-\text{b})=0$
$\Rightarrow\text{a}-\text{b}=0$
View full question & answer→Question 851 Mark
If $x + y = 8$ and $xy = 15,$ than $x^2 + y^2$
Answer$x^2 + y^2 = (x + y)^2 - 2xy$
$\Rightarrow x^2 + y^2 = (8)^2 - 2 \times 15$
$\Rightarrow x^2 + y^2 = 64 - 30$
$\Rightarrow x^2 + y^2 = 34$
View full question & answer→Question 861 Mark
If $\frac{\text{a}}{\text{b}}+\frac{\text{b}}{\text{a}}=1,$ then a3 + b3 =
Answer- 0
Solution:
Here, $\frac{\text{a}}{\text{b}}+\frac{\text{b}}{\text{a}}=1$
$\Rightarrow\frac{\text{a}^2+\text{b}^2}{\text{ab}}=1$
$\Rightarrow\text{a}^2+\text{b}^2=\text{ab}$
$\Rightarrow\text{a}^2+\text{b}^2-\text{ab}=0$
Using, $\text{a}^2+\text{b}^2=(\text{a}+\text{b})(\text{a}^2+\text{b}^2-\text{ab})$
$=(\text{a}+\text{b})(0)$
$=0$
View full question & answer→Question 871 Mark
The coefficient of $x$ in the expansion of $(x + 3)^3$ is:
Answer$(x + 3)^3$
$= x^3 + 3^3 + 9x(x + 3)$
$= x^3 + 27 + 9x^2 + 27x$
So, the coefficient of $x$ in $(x + 3)^3$ is $27.$
View full question & answer→Question 881 Mark
If $a + b + c = 0$ then $\Big(\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}\Big)=?$
Answer$a + b + c = 0$
$\Rightarrow a^3 + b^3 + c^3 = 3abc$
Thus, we have:
$\Big(\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}\Big)=\frac{\text{a}^3+\text{b}^3+\text{c}^3}{\text{abc}}$
$=\frac{3\text{abc}}{\text{abc}}$
$=3$
View full question & answer→Question 891 Mark
The value of $(249)^2 - (248)^2$ is:
Answer$(249)^2 - (248)^2$
$= (249 + 248) (249 - 248) [$Using identity $a^2 - b^2 = (a + b) (a - b)]$
$= 497 \times 1$
$= 497$
View full question & answer→Question 901 Mark
If $a - b = -8$ and $ab = -12,$ then $a^3 - b^3 =$
AnswerUsing,$ (a - b)^3 = a^3 - b^3 - 3ab(a - b)$
$\Rightarrow a^3 - b^3 = (a - b)^3 + 3ab(a - b)$
$\Rightarrow a^3 - b^3 = (-8)^3 + 3(-12) (-8)$
$\Rightarrow a^3 - b^3 = -512 + 288 = -224$
View full question & answer→Question 911 Mark
The zeros of the polynomial $p(x) = x^2 + x - 6$ are.
AnswerThe given polynomial is $p(x) = x^2 + x - 6$
Putting $x = 2$ in $p(x),$ we get
$p(2) = 2^2 + 2 - 6 = 4 + 2 - 6 = 0$
Therefore$, x = 2$ is a zero of the polynomial $p(x).$
Putting $x = -3$ in $p(x),$ we get
$p(-3) = (-3)^2 - 3 - 6 = 9 - 9 = 0$
Therefore$, x = -3$ is a zero of the polynomial $p(x)$
Thus$, 2$ and $-3$ are the zeros of the given polynomial $p(x).$
View full question & answer→Question 921 Mark
If $x - 2$ is a factor of $x^2 + 3ax - 2a,$ then $a =$
AnswerLet $p(x) = x^2 + 3ax - 2a$ be the given polynomial.
$x - 2$ is a factor of $p(x).$
Thus,
$p(2) = 0$
$(2)^2 + 3a \times 2 - 2a = 0$
$4 + 4a = 0$
$a = -1$
View full question & answer→Question 931 Mark
If $x + 2$ and $x - 1$ are the factors of $x^3 + 10x^2 + mx + n,$ then the values of m and $n$ are respectively
AnswerIf $(x + 2)$ and $(x - 1)$ are factors of polynomial $x^3 + 10x^2 + mx + n,$
then $x = -2, x = +1$ will satisfy the polynomial.
Let $p(x) = x^3 + 10x^2 + mx + n$
Then$, p(-2) = 0$
$(-2)^3 + 10(-2)^2 + m(-2) + n = 0$
$-8 + 40 - 2\ m + n = 0$
$32 - 2\ m + n = 0 ...(1)$
And$, p(1) = 0$
$(1)^3 + 10(1)^2 + m(1) + n = 0$
$1 + 10 + m + n = 0$
$11 + m + n = 0 ...(2)$
Substracting equation $(1)$ from equation $(2),$ we get
$-21 + 3\ m = 0$
$3\ m = 21$
$m = 7$
Substituting $m = 7$ in equation $(2),$
$11 + 7 + n = 0$
$18 + n = 0$
$n = -18$
View full question & answer→Question 941 Mark
$(x + 1)$ is a factor of $x^n +1$ only if.
AnswerThe linear polynomial $(x - 1)$ is a factor of $xn + 1,$ only if,
$f(-1) = (-1)^n + 1 = 0$
If n is odd integer$,$ then $f(-1) = -1 + 1 = 0$
View full question & answer→Question 951 Mark
One factor of $x^4 + x^2 - 20$ is $x^2 + 5$. The other factor is:
Answer$x^4 + x^2 - 20$
$= x^4 + 5x^2 - 4x^2 - 20$
$=x^2(x^2 + 5) - 4(x^2 + 5)$
$= (x^2 + 5)(x^2 - 4)$
So$,$ other factor is $x^2 - 4.$
View full question & answer→Question 961 Mark
Which of the following is a binomial?
AnswerA polynomial with two non$-$zero terms is called a binomial.
$x^2 + 4$ is the polynomial that has two non$-$zero terms.
Hence is a binomial.
View full question & answer→Question 971 Mark
A polynomial of degree ____ is called a linear polynomial.
Answer- 1
Solution:
A polynomial of degree 1 is called a linear polynomial.
Its general form is ax + b
View full question & answer→Question 981 Mark
The degree of the polynomial $(x^3 - 2) (x^2 - 11)$ is:
Answer$(x^3 - 2) (x^2 - 11)$
$= x^3(x^2 - 11) - 2(x^2 - 11)$
$= x^5 - 11x^3 - 2x^2 + 22$
Here, the highest power is $5.$
Therefore$,$ the degree is $5.$
View full question & answer→Question 991 Mark
If $(x + 1)$ is a factor of the polynomial $(2x^2 + kx)$ then $k =$ ?
AnswerLet $p(x) = 2x^2 + kx$
Since $(x + 1)$ is a factor of $p(x),$
$= P(-1) = 0$
$\Rightarrow 2(-1)^2 + k(-1) = 0$
$\Rightarrow 2 - k = 0$
$\Rightarrow k = 2$
View full question & answer→Question 1001 Mark
If $x^{51} + 51$ is divided by $x + 1,$ then the remainder is:
Answer$x^{51} + 51$
If $x^{51} + 51$ is divided by $x + 1,$ then using remainder theorem
$(-1)^{51} + 51$
$= -1 + 51$
$= 50$
View full question & answer→