Polynomials Test

Result

Polynomials Test - 47

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Weekly Quiz Competition

Question 1
1 / -0

the values of a& b so that the polynomial $${ x }^{ 3 }-{ ax }^{ 2 }-13x+b\quad is\quad divisible\quad by\quad \left( x-1 \right) \& \left( x+3 \right) $$ are

A
a=15,b=3

B
a=3,b=15

C
a=-3,b=15

D
a=3,b=-15
Question 2
1 / -0

Find the sum of all possible values of $$C$$ for which the expression $${x^3} + 3{x^2} - 9x + c$$ can be expressed as the product of three factors, two of which are identical and monic.

A
$$2$$

B
$$-22$$

C
$$-2$$

D
$$-27$$

Solution
Question 3
1 / -0

The real number 'K' for which teh equation $${\text{2}}{{\text{x}}^3} + 3x + K = 0$$ has to distinct real roots in [0, 1]

A
lies between 1 and 2

B
lies between 2 and 3

C
lies between -1 and 0

D
does not exist
Question 4
1 / -0

If two roots of the equation $$ x ^ { 3 } - 3 x + 2 = 0 $$ are same, then the roots will be

A
2, 2, 3

B
1, 1, -2

C
-2, 3, 3

D
-2, -2, 1

Solution
Question 5
1 / -0

The expression $$(x+(x^4-1)^{1/2})^4 + (x-(x^4-1)^{1/2})^4$$ is a polynomial of degree

A
8

B
6

C
4

D
2

Solution
Question 6
1 / -0

If $$x^5-5qx+4r$$ is divisible by $$(x-c)^2$$ then which of the following must hold true $$\forall q,r,c \in R?$$

A
$$q=r$$

B
$$q+r=0$$

C
$$q^5=r^4$$

D
$$q^4=r^5$$
Question 7
1 / -0

The quadratic polynomials $$f(x) = ax^2 + bx + c $$ has real zeroes $$\alpha$$ and $$\beta$$ . If $$a$$ , $$b$$ , $$c$$ are real and of the same sign then

A
$$\alpha > 0$$ , $$\beta < 0 $$

B
$$\beta > 0$$ , $$\alpha > 0$$

C
$$\alpha < 0$$ , $$\beta > 0 $$

D
$$\beta < 0$$ , $$\alpha < 0$$

Solution
Question 8
1 / -0

If $$\alpha ,\beta ,\gamma$$ are zeroes of the polynomial $${ x }^{ 3 }-x-1$$, then the value of $$\frac { 1+\alpha }{ 1-\alpha } +\frac { 1+\beta }{ 1-\beta } +\frac { 1+\gamma }{ 1-\gamma }$$ is?

A
$$7$$

B
$$-1$$

C
$$-7$$

D
$$1$$

Solution
Question 9
1 / -0

Which of the following is a perfect cube?

A
$$8{x^3} + 36{x^2}y - 54x{y^2} + 27{y^3}$$

B
$${x^3} - 3x - \frac{3}{x} + \frac{1}{{{x^2}}}$$

C
$$8{x^3} - 4\frac{{{x^2}}}{y} + \frac{2}{3}\frac{x}{{{y^2}}} - \frac{1}{{27{y^3}}}$$

D
$$27{x^3} + 54{x^2}y - 36x{y^2} + 8{y^3}$$

Solution

$${\textbf{Step -1: Define a perfect cube}}{\textbf{.}}$$
$${\text{It is a number that is obtained by multiplying the same integer three times}}{\text{.}}$$
$${\textbf{Step -2: Try to simplify the equation}}{\textbf{.}}$$
$${\text{Option A, }}8{x^3} + 36{x^2}y - 54x{y^2} + 27{y^3}.$$
$$ = {\left( {2x} \right)^3} + {\left( {3y} \right)^3} + 18xy\left( {2x - 3y} \right)$$
$${\mathbf{\left[ {{{\left( {a - b} \right)}^3} = {a^3} - {b^3} - 3{a^2}b + 3a{b^2}{\textbf{ and }}{{\left( {a + b} \right)}^3} = {a^3} + {b^3} + 3{a^2}b + 3a{b^2}} \right]}}$$
$$ = {\left( {2x + 3y} \right)^3} + 18xy\left( {2x - 3y} \right)$$
$${\text{Hence, it does not satisfy either of the formula}}{\text{.}}$$
$${\text{Option B, }}{x^3} - 3x - \dfrac{3}{x} + \dfrac{1}{{{x^2}}}.$$
$${\text{As it contains variables with }}\dfrac{1}{{{x^2}}},{\text{ hence not according to formula}}{\text{.}}$$
$${\text{Option C, 8}}{x^3} - 4\dfrac{{{x^2}}}{y} + \dfrac{2}{3}.\dfrac{x}{{{y^2}}} - \dfrac{1}{{27{y^3}}}.$$
$${\text{From the formula,}}$$
$${\left( {2x} \right)^3} - {\left( {\dfrac{1}{{3y}}} \right)^3} - 3 \times {\left( {2x} \right)^2} \times \dfrac{1}{{3y}} + 3 \times \left( {2x} \right) \times {\left( {\dfrac{1}{{3y}}} \right)^2}$$
$${\text{Hence, it fulfills the formula criteria of }}{\left( {a - b} \right)^3}.$$
$${\text{Option D,}}27{x^3} + 54{x^2}y - 36x{y^2} + {\text{ }}8{y^3}.$$
$$ = {\left( {3x} \right)^3} + {\left( {2y} \right)^3} + 18xy\left( {3x - 2y} \right)$$
$$ = {\left( {3x + 2y} \right)^3} $$
$${\textbf{Hence, 8}}{\mathbf{{x^3} - 4\dfrac{{{x^2}}}{y} + \dfrac{2}{3}.\dfrac{x}{{{y^2}}} - \dfrac{1}{{27{y^3}}}}}{\textbf{ is a perfect cube}}.$$
Question 10
1 / -0

The HCF of two ploynomials $$A$$ and $$B$$ using long division method was found to be $$2x + 1$$ after two steps . The fisrt two quotient obtained are $$x$$ and $$(x + 1)$$ . Find $$A$$ and $$B$$ . Given that degree of $$A$$ > degree of $$B$$ is

A
$$A = 2x^3 + 3x^2 + x - 1 $$ ; $$B = 2x^2 - 3x + 1 $$

B
$$A = 2x^3 - 3x^2 + x - 1 $$ ; $$B = 2x^2 - 3x - 1 $$

C
$$A = 2x^3 + 3x^2 - 3x - 1 $$ ; $$B = 2x^2 - 3x + 1 $$

D
$$A = 2x^3 + 3x^2 + 3x + 1 $$ ; $$B = 2x^2 + 3x + 1 $$

Solution