Polynomials Test

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Polynomials Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

If one of the zeroes of a quadratic polynomial of the form $$x^2+ax + b $$ is the negative of the other, then it

A
has no linear term and the constant term is negative.

B
has no linear term and the constant term is positive.

C
can have a linear term but the constant term is negative.

D
can have a linear term but the constant term is positive.

Solution

If one of the zero of the equation is negative of the other, then the roots will be $$m$$ and $$-m$$
Hence, sum of roots $$= m - m = 0$$
Sum of roots $$= -a = 0$$
Thus there will be no linear involved in the term in the equation.
Also, the constant term $$= b =$$ product of the zeroes $$= -m^2$$, will be negative.
Question 2
1 / -0

The coefficient of $$x$$ in the expansion of $$(x + 3)^3$$ is:

A
$$1$$

B
$$9$$

C
$$18$$

D
$$27$$

Solution

Using the formula, $$(a+b)^3 = a^3 +b^3 + 3a^2b + 3ab^2$$
$$(x + 3)^3 = x^3 + 27 + 9x^2 + 27x$$
Therefore, coefficient of $$x$$ is 27
Question 3
1 / -0

Given that one of the zeroes of the cubic polynomial $$ax^3 + bx^2 + cx + d $$ is zero, the product of the other two zeroes is

A
$$-\frac{c}{a}$$

B
$$\frac{c}{a}$$

C
0

D
$$-\frac{b}{a}$$

Solution

For the given equation, $$ax^3 + bx^2 + cx +d$$, 0 is the root of this equation implies d=0
Hence, the equation is $$x(ax^2 + bx +c)$$
The other two roots will be from $$ax^2 +bx +c$$
Product of the other roots = $$\displaystyle \frac{c}{a}$$
Question 4
1 / -0

If $$\alpha,\beta,\gamma$$ are the roots of $$x^{3}+px^{2}+qx+r=0$$, then $$\displaystyle\sum\alpha^2(\beta+\gamma)$$ is

A
$$3r+pq$$

B
$$3r-pq$$

C
$$pq-3r$$

D
$$pq+r$$

Solution

Given: $$\alpha ,\beta ,\gamma $$ are the roots of $$x^{ 3 }+px^{ 2 }+qx+r=0$$

We have
$$\alpha +\beta +\gamma =-p$$
$$\alpha \beta +\beta \gamma +\gamma \alpha =q$$
$$\alpha \beta \gamma =-r$$

Now,
$$\sum { \alpha^2 \left( \beta +\gamma \right) } =(\alpha ^{ 2 }\beta +\alpha^2\gamma) + (\beta^2\gamma+\beta^2\alpha) + (\gamma^2\alpha+\gamma^2\beta)$$

$$=\left( \alpha +\beta +\gamma \right) \left( \alpha \beta +\beta \gamma +\gamma \alpha \right) -3\alpha \beta \gamma$$

$$=-pq+3r$$

$$=3r-pq$$
Question 5
1 / -0

If$$\alpha,\beta and \gamma$$

are the zeroes
of the polynomial$$ x^{3}

5x^{2} 2x + 24$$, which of the following is the difference of two

zeroes?

A
1

B
2

C
3

D
4
Question 6
1 / -0

The coefficient of x in the expansion of $$(x+3)^3$$ is :

A
1

B
9

C
18

D
27
Question 7
1 / -0

Which of the following is/ are NOT a graph of a quadratic polynomial?

A

B

C

D

Solution

(a) Since, the graph meets the x-axis in two distinct points A, B.
$$\therefore$$ it is a graph of a quadratic.
(b) Since, the graph meets the x-axis in two distinct points A, B, but it has taken a turn in the fourth quadrant.
$$\therefore$$ it is not a graph of a quadratic.
(c) Since, the graph meets the x-axis at points A, B (distinct points)
$$\therefore$$ it is a graph of a quadratic.
(d) Since, the graph meets the x-axis at a single point.
$$\therefore$$ it can be a graph of the quadratic equation having two equal real roots.
Question 8
1 / -0

If $$\alpha, \beta, \gamma$$ are the zeroes of the cubic polynomial $$ax^3+bx^2+cx+d=0$$, then $$\alpha+\beta+\gamma=\dfrac {-b}{a}$$.

A
True

B
False

C
Can't be determined

D
None of the above

Solution

We know that if $$\alpha, \beta, \gamma$$ are the zeroes of $$ax^3+bx^2+cx+d$$, then $$\alpha+\beta+\gamma=-\frac {b}{a}, \alpha \beta+\beta \gamma+\gamma \alpha=\frac {c}{a}, \alpha \beta, \gamma=-\frac {d}{a}$$
Question 9
1 / -0

Given that $$x- \sqrt{5}$$ is a factor of the cubic polynomial $$x^3 -3 \sqrt{5}x^2 + 13x -3 \sqrt{5}$$ , find all the zeroes of the polynomial.

A
$$\sqrt{5}, \sqrt{5}+ \sqrt{2}, \sqrt{5} -\sqrt{2}$$

B
$$\sqrt{5}, \sqrt{5}, \sqrt{5}- \sqrt{2}$$

C
$$\sqrt{5}, \sqrt{5} +\sqrt{2}, \sqrt{5} $$

D
None of the above

Solution

If $$(x-\sqrt { 5 })$$ is a factor, then we can write:

$$x^3–3\sqrt { 5 } x^2+13x–3\sqrt { 5 } =(x–\sqrt { 5 } )(x^2+bx+3)$$

To determine the coefficient $$b$$, let's expand the product:

$$(x–\sqrt { 5 } )(x^2+bx+3)=x^3+bx^2+3x–(\sqrt { 5 } )x^2–(\sqrt { 5 } )bx–3\sqrt { 5 }$$

$$(x–\sqrt { 5 } )(x^2+bx+3)=x^3+(b–\sqrt { 5 } )x^2+(3–b\sqrt { 5 } )x–3\sqrt { 5 }$$

Comparing the right hand side to the original expression, we obtain

$$b–\sqrt { 5 } =-3\sqrt { 5 } ⇒b=-2\sqrt { 5 }$$, or, with the same result:

$$3–b\sqrt { 5 } =13\\ ⇒b\sqrt { 5 } =-10\\ ⇒b=-10/\sqrt { 5 } =-2\sqrt { 5 } \\ ⇒b=-2\sqrt { 5 }$$

Therefore,

$$x^3–3\sqrt { 5 } x^2+13x–3\sqrt { 5 } =(x–\sqrt { 5 } )(x^2–2\sqrt { 5 } x+3)\\ x^3–3\sqrt { 5 } x^2+13x–3\sqrt { 5 } =0\\ (x–\sqrt { 5 } )=0,\quad (x^2–2\sqrt { 5 } x+3)=0\\ x–\sqrt { 5 } =0⇒x=\sqrt { 5 } \\ x^2–2\sqrt { 5 } x+3=0⇒x=\sqrt { 5 } ±\sqrt { 2 }$$

Hence, the zeros of the given expression are $$\sqrt { 5 } +\sqrt { 2 } ,\sqrt { 5 } -\sqrt { 2 } ,\sqrt { 5 }$$.
Question 10
1 / -0

Which of the following equations has the sum of its roots as 3?

A
$$2x^2 + 3x + 6 = 0$$

B
$$x^2 - 3x + 3 = 0$$

C
$$\sqrt{2}x^2+\dfrac{3}{\sqrt{2}}x =1 0$$

D
$$3x^2+ 3x + 3 = 0$$

Solution