Question 11 Mark
A polynomial containing two non$-$zero terms is called a $..........$
AnswerA polynomial containing two non$-$zero terms is called a binomial.
$3x + 4, 5y + 1, x^2 + x$
View full question & answer→Question 21 Mark
Which of the following expression is a polynomial.
Answer- $\text{x}^3-1$
Solution:
In a polynomial, the power of the variable of each term should be a whole number.
Here, the power of variable x is 3, which is a whole number.
Therefore, it is a polynomial.
View full question & answer→Question 31 Mark
The factors of $x^2 + 4y^2 + 4y - 4xy - 2x - 8,$ are:
Answer$x^2 + 4y^2 + 4y - 4xy - 2x - 8$
$= x^2 +(2y)^2 - 2 \times x(2y) + 4y - 2x - 8$
$= (x - 2y)^2 + 4y - 2x - 8 ...(1)$
Now making $eq\ (1)$ a perfect square by adding $1$ and $-1$
$(x - 2y)^2 + 4y - 2x - 8 = (x - 2y)^2 + 4y - 2x - 8 + 1 - 1$
$= (x - 2y)^2 + (1)^2 - 2 \times (1) \times (x - 2y) - 9$
$= (x - 2y - 1)^2 - (3)^2$
$= [(x - 2y - 1) - 3][x - 2y - 1 + 3]$
$= (x - 2y - 4)(x - 2y + 2)$
View full question & answer→Question 41 Mark
If $\text{x}+\frac{1}{\text{x}}=2,$ then $\text{x}^3+\frac{1}{\text{x}^3}=$
Answer- 2
Solution:
On cubing we get
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^3=\text{x}^3+\Big(\frac{1}{\text{x}^3}\Big)+3\text{x}\times\frac{1}{\text{x}}\Big(\text{x}+\frac{1}{\text{x}}\Big)$
$\Rightarrow2^3=\text{x}^3+\Big(\frac{1}{\text{x}^3}\Big)+3\times2$
$\Rightarrow\text{3}^3+\Big(\frac{1}{\text{x}^3}\Big)=8-6$
$\Rightarrow\text{x}^3+\Big(\frac{1}{\text{x}^3}\Big)=2$
View full question & answer→Question 51 Mark
$(207 \times 193) =$ ?
Answer$207 \times 193$
$= (200 + 7)(200 - 7)$
$= (200)^2 - (7)^2$
$= 40000 - 49$
$= 39951$
View full question & answer→Question 61 Mark
Which of the following expression is a polynomial in one variable?
Answer- $\sqrt2\text{x}^2-\sqrt3\text{x}+6$
Solution:
Clearly, $\sqrt2\text{x}^2-\sqrt3\text{x}+6$ is a polynomial in one variable because it has only non-negative integral powers of x.
View full question & answer→Question 71 Mark
$x + 1$ is a factor of the polynomial.
Answer$x^3 + 2x^2 - x - 2$
$= x^2\ (x + 2) - 1(x + 2)$
$= (x^2 - 1) (x + 2)$
$= (x + 1) (x - 1) (x + 2)$
View full question & answer→Question 81 Mark
The Expanded of $(3x - 5)^3$ is:
Answer$(3x - 5)^3$
$= (3x)^3 - (5)^3 - 3 \times 3x \times 5(3x - 5)$
$= 27x^3 - 125 - 135x^2 + 225x$
$= 27x^3 - 135x^2 + 225x - 125$
View full question & answer→Question 91 Mark
If $\text{x}^4+\frac{1}{\text{x}^4}=194,$ then $\text{x}^3+\frac{1}{\text{x}^3}=$
Answer$\text{x}^4+\frac{1}{\text{x}^4}=194$
Now $\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^2=\text{x}^4+\frac{1}{\text{x}^4}+2$
$\Rightarrow\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^2=194+2=196$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=14\ ...(1)$
Now $\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\Big\{\text{x}^2+\frac{1}{\text{x}^2}=14\Big\}$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=14+2=16\ [$From $(1)]$
$\Rightarrow\text{x}+\frac{1}{\text{x}}=\sqrt{16}$
$\Rightarrow\text{x}+\frac{1}{\text{x}}=4\ ...(3)$
By identity $a^3 + b^3 = (a + b)(a^2 + b^2 - ab)$
$\Rightarrow\text{x}^3+\frac{1}{\text{x}^3}=\Big(\text{x}+\frac{1}{\text{x}}\Big)\Big(\text{x}^2+\frac{1}{\text{x}^2}-1\Big)$
$=(4)(14-1)$
$=4\times13$
$=52$
View full question & answer→Question 101 Mark
If $\text{x}+\frac{1}{\text{x}}=3,$ then $\text{x}^6+\frac{1}{\text{x}^6}=$
Answer- 322
Solution:
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\text{x}+\frac{1}{\text{x}}=3$ (given)
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=(3)^2-2$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=7\ ...(1)$
Cubing both side of equation (1). we have
$\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^3=(7)^3$
$\Rightarrow(\text{x}^2)^3+\Big(\frac{1}{\text{x}^2}\Big)^3+3(\text{x}^2)\frac{1}{\text{x}^2}\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)=7^3$
$\Rightarrow\text{x}^6+\frac{1}{\text{x}^6}+3(7)=7^3$
$\Rightarrow\text{x}^6+\frac{1}{\text{x}^6}=343-21$
$\Rightarrow\text{x}^6+\frac{1}{\text{x}^6}=322$
Hence, correct option is (d).
View full question & answer→Question 111 Mark
A symbol having a fixed value is called a ______.
Answer- Constant.
Solution:
A symbol of having a fixed value is called a constant.
Ex. Any natural number, whole number, integers, rational number.
A symbol having the variable values is called variable.
View full question & answer→Question 121 Mark
If $\text{x}+\frac{1}{\text{x}}=4,$ then $\text{x}^4+\frac{1}{\text{x}^4}=$
Answer- 194
Solution:
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)=4$ (given)
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=(4)^2-2=16-2=14\ ...(1)$
Squaring equation (1)
$\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^2=(14)^2$
$\Rightarrow(\text{x}^2)^2+\Big(\frac{1}{\text{x}^2}\Big)^2+2(\text{x}^2)\frac{1}{\text{x}^2}=196$
$\Rightarrow\text{x}^4+\frac{1}{\text{x}^4}=196-2$
$\Rightarrow\text{x}^4+\frac{1}{\text{x}^4}=194$
Hence, correct option is (b).
View full question & answer→Question 131 Mark
Write the correct answer in the following: The value of the polynomial $5x - 4x^2 + 3,$ when $x = -1$ is.
AnswerLet $p(x) = 5x - 4x^2 + 3 ... (i)$
On putting $x= -1$ in $eq. (i),$ we get
$p(-1) = 5(-1) - 4(-1)^2 + 3 = -5 - 4 + 3 = -6$
View full question & answer→Question 141 Mark
If $\text{p}(\text{x})=\text{x}^2-2\sqrt2\text{x}+1$ then $\text{p}(2\sqrt2)=?$
Answer- 1
Solution:
$\text{p}(\text{x})=\text{x}^2-2\sqrt2\text{x}+1$
$\text{p}(2\sqrt2)=(2\sqrt2)^2-2\sqrt2(2\sqrt2)+1$
$=8-8+1$
$=1$
View full question & answer→Question 151 Mark
Write the correct answer in the following:
If $\frac{\text{x}}{\text{y}}+\frac{\text{y}}{\text{x}}=-1 \ (\text{x},\text{y}\neq0),$ the value of $\text{x}^3-\text{y}^3$ is.
Answer- 0
Solution:
Given, $\frac{\text{x}}{\text{y}}+\frac{\text{y}}{\text{x}}=-1$
$\Rightarrow\frac{\text{x}^2+\text{y}^2}{\text{xy}}=-1$
$\Rightarrow\text{x}^2+\text{y}^2=-\text{xy}$
$\Rightarrow\text{x}^2+\text{y}^2+\text{xy}=0$
Now, $\text{x}^3-\text{y}^3=(\text{x}-\text{y})(\text{x}^2+\text{xy}+\text{y}^2) \ ...(\text{i})$
$[\text{a}^3-\text{b}^3=(\text{a}-\text{b})(\text{a}^2+\text{ab}+\text{b}^2)]$
$=(\text{x}-\text{y})\times0=0$ [From Eq. (i)]
View full question & answer→Question 161 Mark
If $a + b + c = 0$ then $(a^3 + b^3 + c^3)$ is:
Answer$a + b + c = 0$
$\Rightarrow a + b = -c$
$\Rightarrow (a + b)^3 = (-c)^3$
$\Rightarrow a^3 + b^3 + 3ab(a + b) = -c^3$
$\Rightarrow a^3 + b^3 + 3ab(-c) = -c^3$
$\Rightarrow a^3 + b^3 + c^3 = 3abc$
View full question & answer→Question 171 Mark
If$ a + b + c = 9$ and $ab + bc + ca = 23,$ then $a^2 + b^2 + c^2 =$
AnswerWe know that$ (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$
Here$, a + b + c = 9, ab + bc + ca = 23$
Thus, we have
$(9)^2 = a^2 + b^2 + c^2 + 2(23)$
$81 = a^2 + b^2 + c^2 + 46$
$a^2 + b^2 + c^2 = 81 - 46$
$a^2 + b^2 + c^2 = 35$
View full question & answer→Question 181 Mark
Which of the following expression is a polynomial?
AnswerA polynomial is an algebraic expression in which the variables involved have only non$-$negative integrals powers.
Option $(a), (b)$ and $(c)$ have negative and non$-$integral powers,
So they are not polynomials.
In option $(d),$
$\text{x}^2+\frac{2\text{x}^\frac{3}{2}}{\sqrt{\text{x}}}+6=\text{x}^2+2\text{x}^{\frac{3}{2}-\frac{1}{2}}+6$
$x^2 + 2x^1 + 6$ Which is a polynomial.
View full question & answer→Question 191 Mark
If $(x^{100} + 2x^{99} + k)$ is divisible By $(x + 1)$ then the value of $k$ is:
AnswerLet$: (x^{100} + 2x^{99} + k)$
Now$, x + 1 = 0$
$\Rightarrow x = -1$
$\therefore p(-1) = 0$
$\Rightarrow (1)^{100} + 2 \times (-1)^{99} + k = 0$
$\Rightarrow 1 - 2 + k = 0$
$\Rightarrow -1 + k = 0$
$\Rightarrow k = 1$
View full question & answer→Question 201 Mark
When $x^3 - 2x^2 + ax - b$ is divided by$ x^2 - 2x - 3,$ the remainder is $x - 6.$ The values of $a$ and $b$ are respectively
AnswerLet $p(x) = x^3 - 2x^2 + ax - b, r(x) = x - 6$ and $q(x) = x^2 - 2x - 3$
Then $q(x)$ is a factor of $[p(x) - r(x)] [$because if $p(x)$ is divided by $q(x),$ remainder is $r(x)].$
So $, [p(x) - r(x)]$ will be exactly divided by $q(x)]$
Now,
$q(x) = x^2 - 2x - 3 = (x - 3)(x + 1)$
If $q(x)$ is a factor of$ [p(x) - r(x)]$
Then $(x - 3)$ and $(x + 1)$ are also factors of$ [p(x) - r(x)]$
So, at $x = 3$ and $x = -1, p(x) - r(x)$ will be zero.
Now
$p(3) - r(3) = 0$
$i.e. (3)^3 - 2(3)^2 + a(3) - b - (3 - 6) = 0$
$i.e. 27 - 18 + 3a - b + 3 = 0$
$i.e. 3a - b + 12 = 0 ...(1)$
And,
$p(-1) - r(-1) = 0$
$i.e. (-1)^3 - 2(-1)^2 + a(-1) - b - (-1 - 6) = 0$
$i.e. -1 - 2 - a - b + 7 = 0$
$i.e -a - b + 4 = 0 ...(2)$
Subtracting equation $(2)$ from equation $(1),$ we get
$4a + 8 = 0$
$a = -2$
From $(2), -(-2) - b + 4 = 0$
$b = 6$
View full question & answer→Question 211 Mark
If $\text{x}+\frac{1}{\text{x}}=5,$ then $\text{x}^2+\frac{1}{\text{x}^2}=$
Answer- 23
Solution:
Using, $(\text{a}+\text{b})^2=\text{a}^2+\text{b}^2+2\text{ab}$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\Big(\frac{1}{\text{x}^2}\Big)+2\text{x}\frac{1}{\text{x}}$
$\Rightarrow(5)^2=\text{x}^2+\Big(\frac{1}{\text{x}^2}\Big)+2$
$\Rightarrow\text{x}^2+\Big(\frac{1}{\text{x}^2}\Big)=25-2$
$\Rightarrow\text{x}^2+\Big(\frac{1}{\text{x}^2}\Big)=23$
View full question & answer→Question 221 Mark
If $\text{x}+\frac{1}{\text{x}}=3,$ than the value of $\text{x}^2+\frac{1}{\text{x}^2}$ is:
Answer- 7
Solution:
$\text{x}+\frac{1}{\text{x}}=3,$
Squaring both sides, we get
$\text{x}^2+\frac{1}{\text{x}^2}+2\times\text{x}\times\frac{1}{\text{x}}=9$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}+2=9$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=7$
View full question & answer→Question 231 Mark
Write the correct answer in the following: Degree of the polynomial $4x^4 + 0x^3 + 0x^5 + 5x + 7$ is.
AnswerThe height power of the variable in a polynomial is called the degree of the polynomial. In this polynomial, the term with highest power of x is $4x^4$. Highest power of x is $4$, so the degree of the given polynomial is $4.$
View full question & answer→Question 241 Mark
The expression $x^4 + 4$ can be factorized as $:$
AnswerThe given expression to be factorized is $x^4 + 4$
This can be written in the form
$x^4 + 4 = (x^2)^2 + (2)^2 + 4x^2 - 4x^2$
$= {(x^2)^2 + 2 \times x^2 \times 2 + (2)^2} - (2x)^2$
$= (x^2 + 2)^2 - (2x)^2$
$= (x^2 + 2 + 2x) (x^2 + 2 - 2x)$
View full question & answer→Question 251 Mark
The value of $(\sqrt{\text{x}}+\sqrt{\text{y}})(\sqrt{\text{x}}-\sqrt{\text{y}})(\text{x}+\text{y})(\text{x}^2+\text{y}^2)$ is:
Answer- $(\text{x}^4-\text{y}^4)$
Solution:
$(\sqrt{\text{x}}+\sqrt{\text{y}})(\sqrt{\text{x}}-\sqrt{\text{y}})(\text{x}+\text{y})(\text{x}^2+\text{y}^2)$
$[(\sqrt{\text{x}})^2-(\sqrt{\text{y}})^2](\text{x}+\text{y})(\text{x}^2+\text{y}^2)$
$=(\text{x}-\text{y})(\text{x}+\text{y})(\text{x}^2+\text{y}^2)$
$=[(\text{x})^2-(\text{y})^2](\text{x}^2+\text{y}^2)$
$=(\text{x}^2-\text{y}^2)(\text{x}^2+\text{y}^2)$
$=(\text{x}^2)^2-(\text{y}^2)^2$
$=\text{x}^2-\text{y}^2$
View full question & answer→Question 261 Mark
If $x^2 - 1$ is a factor of $ax^4 +bx ^3 + cx^2 + dx + e,$ then.
AnswerAs$ (x^2 - 1)$ Is a factor of polynomial
$f(x^2) = ax^4 + bx^3 + cx^2 + dx + e$
Therefore$,$
$f(x) = 0$
And,
$f(1) = 0$
$a(1)^4 + b(1)^3 + c(1)^2 + d(1) + e = 0$
$\Rightarrow a + b + c + d + e = 0$
And,
$f(-1) = 0$
$a(-1)^4 + b(-1)^3 + c(-1)^2 + d(-1) + e = 0$
$a - b - c - d + e = 0$
Hence$, a + c + e = b + d$
View full question & answer→Question 271 Mark
The remainder obtained when the polynomial p(x) is divided by (b - ax) is:
Answer- $\text{p}\Big(\frac{\text{b}}{\text{a}}\Big)$
Solution:
Given: Divisor = b - ax
For getting the remainder we have to find value of x, which is put in p(x).
Then, b - ax = 0
$\Rightarrow\text{x}=\frac{\text{b}}{\text{a}}$
Therefore, the required remainder is $\text{p}(\text{x})=\text{p}\Big(\frac{\text{b}}{\text{a}}\Big)$
View full question & answer→Question 281 Mark
The zeros of the polynomial $\text{p(x)} = 3\text{x}^2 - 1$ are.
Answer- $\frac{1}{\sqrt{3}}\text{ and }\frac{-1}{\sqrt{3}}$
Solution:
Let: $\text{p(x)} = 3\text{x}^2 - 1$
To find the zeroes of p(x),
We have:
$\text{p(x)}=0\Rightarrow3\text{x}^2-1=0$
$\Rightarrow3\text{x}^2=1$
$\Rightarrow\text{x}^2=\frac{1}{3}$
$\Rightarrow\text{x}=\pm\frac{1}{\sqrt{3}}$
$\Rightarrow\text{x}=\frac{1}{\sqrt{3}}\text{ and }\text{x}=\frac{-1}{\sqrt{3}}$
View full question & answer→Question 291 Mark
Zero of the polynomial p(x) = 2x + 5 is:
Answer- $\frac{-5}{2}$
Solution:
The zero of the polynomial p(x) can be obtained by putting p(x)
p(x) = 0
⇒ 2x + 5 = 0
⇒ 2x = -5
$\Rightarrow\text{x}=\frac{-5}{2}$
View full question & answer→Question 301 Mark
The factorization of $9x^2 - 3x - 20$ is:
Answer$9x^2 - 3x - 20$
$= 9x^2 - 15x + 12x - 20$
$= 3x(3x - 5) + 4(3x - 5)$
$= (3x + 4) (3x - 5)$
View full question & answer→Question 311 Mark
The coefficient of $'x\ '$ in the expansion of $(x + 3)^3$ is:
Answer$(x + 3)^3 $
$= x^3 + (3)^3 + 3 \times x \times 3 (x + 3) $
$= x^3 + 27 + 9x^2 + 27x $
$= x^3 + 9x^2 + 27x + 27$
Therefore, the coefficient of$ x$. in the expansion of $(x + 3)^3$ is $27.$
View full question & answer→Question 321 Mark
If $x - a$ is a factor of $x^3 - 3x^2a + 2a^2x + b,$ then the value of $b$ is:
AnswerLet $p(x) = x^3 - 3x^2a + 2a^2x + b$
$(x - a)$ is a factor of $p(x).$
So$,$
$p(a) = 0$
$a^3 - 3a^2a + 2a^2a + b = 0$
$a^3 - 3a^3 + 2a^3 + b = 0$
$3a^3 - 3a^3 + b = 0$
$b = 0$
View full question & answer→Question 331 Mark
For what value of $k$ is the polynomial $p(x) = 2x^3 - kx^2 + 3x + 10$ exactly divisible by $(x + 2)$?
Answer$p(x) = 2x^3 - kx^2 + 3x + 10$
$x + 2 = 0$
$\Rightarrow x = -2$
By the factor theorem, we know that when $p(x)$ is divided by $(x + 2),$ the remainder is $p(-2).$
Now$, p(-2) = 2(-2)^3 + k(-2)^2+ 3(-2) + 10$
$\Rightarrow 0 = -16 - 4k - 6 + 10$
$\Rightarrow 0 = -12 - 4k$
$\Rightarrow 4k = -12$
$\Rightarrow k = -3$
View full question & answer→Question 341 Mark
$(4x^2 + 4x - 3) =$ ?
Answer$(4x^2) + (4x - 3) $
$= 4x^2 + 6x - 2x - 3$
$= 2x(2x + 3) - 1(2x + 3)$
$= (2x + 3) (2x - 1)$
View full question & answer→Question 351 Mark
If $\frac{\text{x}}{\text{y}}+\frac{\text{y}}{\text{x}}=-1(\text{x},\text{y}\not=0),$ the value of $x^3 - y^3$ is:
AnswerGiven: $\frac{\text{x}}{\text{y}}+\frac{\text{y}}{\text{x}}=-1$
$\Rightarrow\frac{\text{x}^2+\text{y}^2}{\text{xy}}=-1$
$\Rightarrow\text{x}^2+\text{y}^2=-\text{xy}\ ...(\text{i})$
Now, $\text{x}^3-\text{y}^3=(\text{x}^2+\text{y}^2+\text{xy})$
$\Rightarrow\text{x}^3-\text{y}^3=(\text{x}-\text{y})(\text{-xy}+\text{xy})\ \ \ [\text{from eq. (i)}]$
$\Rightarrow\text{x}^3-\text{y}^3=(\text{x}-\text{y})(0)$
$\Rightarrow(\text{x}^3-\text{y}^3)=0$
View full question & answer→Question 361 Mark
If $\text{x}+\frac{1}{\text{x}}=3,$ then $\text{x}^6+\frac{1}{\text{x}^6}=$
Answer- 322
Solution:
On cubing we get.
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^3=\text{x}^3+\Big(\frac{1}{\text{x}^3}\Big)+3\times\text{x}\times\frac{1}{\text{x}}\Big(\text{x}+\frac{1}{\text{x}}\Big)$
$\Rightarrow27=\text{x}^3+\Big(\frac{1}{\text{x}^3}\Big)+3\times\text{3}$
$\Rightarrow\text{x}^3+\Big(\frac{1}{\text{x}^3}\Big)=27-9$
$\Rightarrow\text{x}^3+\Big(\frac{1}{\text{x}^3}\Big)=18$
Now, $\Big(\text{x}^3+\frac{1}{\text{x}^3}\Big)^2=\text{x}^6+\Big(\frac{1}{\text{x}^6}\Big)+2\times\text{x}^3\times\frac{1}{\text{x}^3}$
$\Rightarrow18^2=\text{x}^6+\Big(\frac{1}{\text{x}^6}\Big)+2$
$\text{x}^6+\Big(\frac{1}{\text{x}^6}\Big)=324-2=322$
View full question & answer→Question 371 Mark
$6x^2 + 17x + 5 =$ ?
Answer$6x^2 + 17x + 5$
$= 6x^2 + 15x + 2x + 5$
$= 3x(2x + 5) + 1(2x + 5)$
$= (2x + 5)(3x + 1)$
View full question & answer→Question 381 Mark
The zeros of the polynomial $p(x) = 2x^2 + 7x - 4$ are:
Answer$p(x) = 2x^2 + 7x - 4$
Now$, p(x) = 0$
$\Rightarrow 2x^2 + 7x - 4 = 0$
$\Rightarrow 2x^2 + 8x - x - 4 = 0$
$\Rightarrow 2x(x + 4) - 1(x + 4) = 0$
$\Rightarrow (x + 4)(2x - 1) = 0$
$\Rightarrow x + 4 = 0$ and $2x - 1 = 0$
$\Rightarrow x = -4$ and $\text{x}=\frac{1}{2}$
View full question & answer→Question 391 Mark
If $x + y + z = 9$ and $xy + yz + zx = 23,$ the value of $(x^3 + y^3 + z^3 - 3xyz) =$ ?
Answer$x^3 + y^3 + z^3 - 3xyz $
$= (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$
$= (x + y + z)[(x + y + z)^2 - 3(xy + yz + zx)]$
$= 9 \times (81 - 3 \times 23)$
$= 9 \times 12$
$= 108$
View full question & answer→Question 401 Mark
If $x + y + z = 9$ and $xy + zx = 23,$ then the value of $x^3 + y^3 + z^3 - 3xyz$ is:
AnswerGiven: $x + y + z = 9$ and $xy + zx = 23$
$x^3 + y^3 + z^3 - 3xyz = (x + y + z) (x^2 + y^2 + z^2 - xy - yz - zx)$
$= (x + y + z) [(x + y + z)^2 - 2xy - 2yz - 2zx - xy - yz - zx]$
$= (x + y + z) [(x + y + z)^2 - 3xy - 3yz - 3zx]$
$= (x + y + z) [(x + y + z)^2 -3(xy + yz + zx)]$
$= 9 \times [81 - 69]$
$= 9 \times 12$
$= 108$
View full question & answer→Question 411 Mark
If $(x + 2)$ and$ (x - 1)$ are factors of the polynomial $p(x) = x^3 + 10x^2 + mx + n$ then:
AnswerLet $f(x) = x^3 + 10x^2 + mx + n$
Now$, x + 2 = 0$
$\Rightarrow x = -2$
and $x - 1 = 0$
$\Rightarrow x = 1$
By factor theorem,
$f(-2) = 0$
$\Rightarrow (-2)^3 + 10(-2)^2 + m(-2) + n$
$\Rightarrow -8 + 40 - 2m + n = 0$
$\Rightarrow 2m - n = 32 ...(i)$
By factor theorem,
$f(1) = 0$
$\Rightarrow (1)^3 + 10(1)^2 + m(1) + n = 0$
$\Rightarrow m + n = -11 ...(ii)$
Adding $(i)$ and $(ii),$ we get
$3m = 21$
$\Rightarrow m = 7$
Substituting in $(ii),$ we get
$n = -18$
View full question & answer→Question 421 Mark
If $a + b + c = 0,$ then $a^3 + b^3 + c^3$ is equal to:
Answer$If\ a + b + c = 0,$ then
$a^3 + b^3 + c^3 - 3abc = 0$
$\Rightarrow a^3 + b^3 + c^3 = 3abc$
View full question & answer→Question 431 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 441 Mark
$(x^2 - 4x - 21) =$ ?
Answer$(x^2 - 4x - 21) $
$= x^2 - 7x + 3x - 21$
$= x(x - 7) + 3(x - 7)$
$= (x - 7) (x + 3)$
View full question & answer→Question 451 Mark
$(x + y)^3 - (x - y)^3$ can be factorized as:
AnswerPut $a = x + y$ and $b = x - y,$ then
$(x + y)^3 - (x - y)^3 = a^3 - b^3$
$= (a − b) (a^2 + b^2 + ab)$
$= (x + y - x + y) [(x + y)^2 + (x - y)^2 + (x - y) (x+y)]$
$=2y[2(x^2 + y^2) + (x^2 - y^2)]$
$=2y[3x^2 + y^2]$
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$\frac{(\text{a}^2-\text{b}^2)^3+(\text{b}^2-\text{c}^2)^3+(\text{c}^2-\text{a}^2)^3}{(\text{a}-\text{b})^3+(\text{b}-\text{c})^3+(\text{c}-\text{a})^3}=$
AnswerIf $a + b + c = 0$ then, $a^3 + b^3 + c^3 = 3abc$
Now$, (a^2 - b^2) + (b^2 - c^2) + (c^2 - a^2) = a^2 - b^2 + b^2 - c^2 + c^2 - a^2 = 0$
$\Rightarrow (a^2 - b^2)^3 + (b^2 - c^2)^3 + (c^2 - a^2)^3 = 3(a^2 - b^2)(b^2 - c^2)(c^2 - a^2)$
Again$, (a - b) + (b - c) + (c - a) = a - b + b - c + c - a = 0$
$\Rightarrow (a - b)^3 + (b - c)^3 + (c - a)^3 = 3(a - b)(b - c)(c - a)$
Thus, we have
$\frac{(\text{a}^2-\text{b}^2)^3+(\text{b}^2-\text{c}^2)^3+(\text{c}^2-\text{a}^2)^3}{(\text{a}-\text{b})^3+(\text{b}-\text{c})^3+(\text{c}-\text{a})^3}$
$=\frac{3(\text{a}^2-\text{b}^2)(\text{b}^2-\text{c}^2)(\text{c}^2-\text{a}^2)}{3(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})}$
$=\frac{(\text{a}-\text{b})(\text{a}+\text{b})(\text{b}-\text{c})(\text{b}+\text{c})(\text{c}-\text{a})(\text{c}+\text{a})}{(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})}$
$=(\text{a}+\text{b})(\text{b}+\text{c})(\text{c}+\text{a})$
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When $p(x) = 4x^3 - 12x^2 + 11x - 5$ is divided by $(2x - 1),$ the remainder is:
Answer$\text{p}(\text{x}) = 4\text{x}^3 - 12\text{x}^2 + 11\text{x} - 5$
$\text{x}-1=0$
$\Rightarrow\text{x}=\frac{1}{2}$
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, $\text{p}\Big(\frac{1}{2}\Big)= 4\text{x}^3 - 12\text{x}^2 + 11\text{x} - 5$
$=4\Big(\frac{1}{2}\Big)^3-12\Big(\frac{1}{2}\Big)^2+11\Big(\frac{1}{2}\Big)-5$
$=\frac{1}{2}-3+\frac{11}{2}-5$
$=-2$
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$\sqrt{2}$ is a polynomial of degree.
Answer- 0
Solution:
$\sqrt{2}$ is a constant term. Therefore, the degree of $\sqrt{2}$ is 0.
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Write the correct answer in the following:
$\sqrt{2}$ is a polynomial of degree.
Answer- 0
Solution:
$\sqrt{2}$ is a constant polynomial. The only term here is $\sqrt{2}$ which can be written as $\sqrt{2}\text{x}^\circ.$ So, the exponent of x is zero. Therefore, the degree of the polynomial is 0.
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Which of the following is a polynomial in one variable?
Answer- $\sqrt{2}\text{x}^2-\sqrt{3}\text{x}+6$
Solution:
Clearly, $\sqrt{2}\text{x}^2-\sqrt{3}\text{x}+6$ is a polynomial in one variable because it has only non-negative integral powers of x.
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