Polynomials Test

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Polynomials Test - 36

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Question 1
1 / -0

If one of the zeros 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

We know that equation $$ax^{2}+bx+c=0$$
Then sum of roots $$=\dfrac{-b}{a}$$ and product of roots$$=\dfrac{c}{a}$$
Given quadratic polynomial is,
$$p(x)=x^2+ax+b$$
Let one of the zero of the polynomial is $$\alpha$$ then the other zero will be $$-\alpha$$.
Sum of zeros=$$\alpha+(-\alpha)=\dfrac { -a }{ 1 }$$

=>$$0=\dfrac { -a }{ 1 }$$

=>$$a=0$$
and product of the zeros =$$\alpha(-\alpha) =\dfrac {b}{1}$$

=>$$-\alpha^2=b$$

=>$$b=-\alpha^2$$
=>$$b$$ has to be negative.
Hence, option $$A$$ is correct.
Question 2
1 / -0

If 2 and $$-\dfrac {1}{2}$$ is the sum and product of the zeros of the polynomials respectively then the quadratic polynomial f(x) is:

A
$$x^2-2x-4$$

B
$$4x^2-2x+1$$

C
$$2x^2+4x-1$$

D
$$2x^2-4x-1$$

Solution

Let say $$\alpha$$, $$\beta$$ be the zeros of the quadratic polynomial $$a{ x }^{ 2 }+bx+c$$ , then

$$\left( \alpha +\beta \right) =2=-\dfrac { b }{ a } ,\quad \alpha \beta =-\dfrac { 1 }{ 2 }=\dfrac { c }{ a } $$

If $$a=2$$, then $$b=-4$$, and $$c=-1$$
Therefore the required polynomial will be $$ 2{ x }^{ 2 }-4x-1$$.
Question 3
1 / -0

Which out of the following options is a trinomial, having degree 7?

A
$$x^7 + 6x - 8$$

B
$$5x^2 + 12x^{-7} + x$$

C
$$y^3 - 4x^2 + 12xy$$

D
$$y^7 - y^6 + 5y^4 + 6 - 8y^5 + \sqrt y$$

Solution

The polynomial $$x^7+6x-8$$ has three terms. The first one is $$x^7$$, the second is $$6x$$, and the third is $$-8$$.

The exponent of the first term is $$7$$.

The exponent of the second term is $$1$$ because $$6x = 6x^1$$.

The exponent of the third term is $$0$$ because $$-8 = -8x^0$$.

Since the highest exponent is $$7$$, therefore, the degree of $$x^7+6x-8$$ is $$7$$.
Question 4
1 / -0

If one of the zeros of the quadratic polynomial $$2x^2 + px + 4$$ is 2, find the other zero. Also find the value of p.

A
$$p=-1$$, other zero $$=1$$

B
$$p=-2$$, other zero $$=1$$

C
$$p=-6$$, other zero $$=1$$

D
$$p=-7$$, other zero $$=1$$

Solution

We know that for a quadratic polynomial $$ax^2+bx+c$$,
product of the zeros=$$\dfrac{c}{a}$$
Here,quadratic polynomial is $$2x^2+px+4$$
One root is $$2$$
Let the other root be $$a$$
Therefore,
$$2\times a=\dfrac{4}{2}$$
$$=>a=1$$
Therefore,
The other zero is $$1$$
Also, sum of the zeros =$$\dfrac{-b}{a}$$
$$=>1+2=\dfrac{-p}{2}$$
$$=>3\times 2=-p$$
$$=>p=-6$$
Therefore,
$$p=-6$$ , other root$$=1$$
Question 5
1 / -0

Given that the zeros of the cubic polynomial $$x^3-6x^2+3x+10$$ are of the form $$a, a + b, a + 2b$$ for some real numbers $$a$$ and $$b$$, find the values of $$a$$ and $$b$$.

A
$$a=4$$ and $$b=2$$ or $$a=3, b=-7$$

B
$$a=5$$ and $$b=-3$$ or $$a=1, b=-7$$

C
$$a=-1$$ and $$b=3$$ or $$a=5, b=-3$$

D
$$a=2$$ and $$b=-6$$ or $$a=-3, b=1$$
Question 6
1 / -0

For what value of $$k$$, $$-2$$ is a zero of the polynomial $$3x^2 + 4x + 2k$$?

A
$$5$$

B
$$3$$

C
$$-8$$

D
$$-2$$

Solution

If $$-2$$ is a zero of $$f(x)=3x^2+4x+2k$$
Thus,
$$f(-2)=0$$
$$\therefore 3(-2)^2+4(-2)+2k=0$$
$$=>12-8+2k=0$$
$$=>2k=-4$$
$$=>k=-2$$
Question 7
1 / -0

Find a cubic polynomial with the sum of its zeros, sum of the product of its zeros taken two at a time and product of its zeros as 2, -7, - 14 respectively.

A
$$k(x^3-2x^2-7x+14)$$

B
$$k(x^3-7x^2-2x+14)$$

C
$$k(x^3+2x^2-7x-14)$$

D
$$k(x^3-7x^2+2x-14)$$

Solution

Let, $$\alpha,\beta,\gamma$$ be the zeros of the given cubic polynomial. Then, we have
$$\alpha+\beta+\gamma=2$$
$$\alpha\beta+\beta\gamma+\gamma\alpha=-7$$
$$\alpha\beta\gamma=-14$$
Now, the required polynomial$$=k\times[x^3-(\alpha+\beta+\gamma)x^2+(\alpha\beta+\beta\gamma+\gamma\alpha)x-(\alpha\beta\gamma)]$$, where k is real constant.
$$=k\times[x^3-2x^2+(-7)x-(-14)]$$
$$=k\times(x^3-2x^2-7x+14)$$
Question 8
1 / -0

The degree of trinomial $$\displaystyle ax^{5}-bx^{4}+c$$ is

A
$$9$$

B
$$5$$

C
$$4$$

D
$$1$$

Solution

The polynomial $$ax^5-bx^4+c$$ has three terms. The first one is $$ax^5$$, the second is $$-bx$$, and the third is $$c$$.

The exponent of the first term is $$5$$.

The exponent of the second term is $$1$$ because $$-bx = -bx^1$$.

The exponent of the third term is $$0$$ because $$c = cx^0$$.

Since the highest exponent is $$5$$, therefore, the degree of $$ax^5-bx^4+c$$ is $$5$$.
Question 9
1 / -0

Draw the graph of the polynomial $$f(x) = 2x -5$$. Also, find the coordinates of the point where it crosses X-axis.

A
$$\dfrac{2}{5}$$

B
$$\dfrac{5}{2}$$

C
$$0$$

D
None of these

Solution

Let $$y =2x-5$$.
The following lists the values of $$y$$ corresponding to different values of $$x$$.
When $$x = 1$$, then $$y = 2(1) - 5 = -3$$
When $$x = 4$$ then $$y = 2(4) - 5 = -3$$

The points $$A (1, -3)$$ and $$B (4, 3)$$ are plotted on the graph paper on a suitable scale. A line is drawn passing through these points to obtain the graphs of the given polynomial

It crosses X-axis at point $$(2.5,0)$$.
Question 10
1 / -0

The number of terms in the quadratic polynomial $$x^2+4x+4+5$$.

A
$$2$$

B
$$4$$

C
$$3$$

D
None of the above

Solution

The quadratic polynomial $$x^2+4x+4+5$$ can be rewritten as $$x^2+4x+9$$.
The first term of the polynomial is $$x^2$$, the second term of the polynomial is $$4x$$ and the third term of the polynomial is $$9$$.
Hence, the number of terms in the given polynomial is $$3$$.

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		The points $$A (1, -3)$$ and $$B (4, 3)$$ are plotted on the graph paper on a suitable scale. A line is drawn passing through these points to obtain the graphs of the given polynomial

Polynomials Test - 36

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Polynomials Test - 36

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Question 1

has no linear term and the constant term is negative

has no linear term and the constant term is positive

can have a linear term but the constant term is negative

can have a linear term but the constant term is positive

Solution

Question 2

$$x^2-2x-4$$

$$4x^2-2x+1$$

$$2x^2+4x-1$$

$$2x^2-4x-1$$

Solution

Question 3

$$x^7 + 6x - 8$$

$$5x^2 + 12x^{-7} + x$$

$$y^3 - 4x^2 + 12xy$$

$$y^7 - y^6 + 5y^4 + 6 - 8y^5 + \sqrt y$$

Solution

Question 4

$$p=-1$$, other zero $$=1$$

$$p=-2$$, other zero $$=1$$

$$p=-6$$, other zero $$=1$$

$$p=-7$$, other zero $$=1$$

Solution

Question 5

$$a=4$$ and $$b=2$$ or $$a=3, b=-7$$

$$a=5$$ and $$b=-3$$ or $$a=1, b=-7$$

$$a=-1$$ and $$b=3$$ or $$a=5, b=-3$$

$$a=2$$ and $$b=-6$$ or $$a=-3, b=1$$

Question 6

$$5$$

$$3$$

$$-8$$

$$-2$$

Solution

Question 7

$$k(x^3-2x^2-7x+14)$$

$$k(x^3-7x^2-2x+14)$$

$$k(x^3+2x^2-7x-14)$$

$$k(x^3-7x^2+2x-14)$$

Solution

Question 8

$$9$$

$$5$$

$$4$$

$$1$$

Solution

Question 9

$$\dfrac{2}{5}$$

$$\dfrac{5}{2}$$

$$0$$

None of these

Solution

Question 10

$$2$$

$$4$$

$$3$$

None of the above

Solution

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