Polynomials Test

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Polynomials Test - 42

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Weekly Quiz Competition

Question 1
1 / -0

Identify the ordered pair of the zeroes of the linear polynomials given in the graph?

A
$$(2, -2)$$ and $$(5,0)$$

B
$$(14, 12)$$ and $$(5,0)$$

C
$$(1.8, 0)$$ and $$(11,0)$$

D
$$(1, 5)$$ and $$(1.8,0)$$

Solution

The graph intersect the $$x$$-axis at $$(11, 0)$$.
Hence $$(11, 0)$$ is the ordered pair of the zeroes of the linear polynomial from the graph as it intersects at $$x$$-axis at $$(11, 0)$$.
Question 2
1 / -0

Which graph is an example of zeroes of the quadratic polynomial?

A

B

C

D

Solution

A quadratic polynomial should cut the x-axis at maximum two points. So, third graph x-coordinates of $$(0, 0)$$ and $$(10, 0)$$ are the two zeroes of the quadratic polynomial.
Question 3
1 / -0

Which graph is an example of zeroes of the cubic polynomial?

A

B

C

D

Solution

The first graph cuts x-coordinates at $$(-3, 0) (-2, 0)$$ and $$(-1, 0)$$ and are the three zeroes of the cubic polynomial.
Question 4
1 / -0

Find the zeroes of the cubic polynomial from the graph.

A
$$1, 2, 3.1$$

B
$$2, -1, 3.1$$

C
$$1, -2, 3.1$$

D
$$-1, -2, -3.1$$

Solution

Zeroes of a polynomial are the values of $$x$$ co-ordinates where its curve intersects $$x$$-axis.
Here $$x$$ co-ordinates of the points where curve intersects $$x$$-axis are $$1, 2, 3.1$$.
Thus zeroes of the given polynomial are $$1,2,3.1$$
Question 5
1 / -0

Identify the zeroes of cubic polynomial $$(x-2)(x-4)(x-6)$$

A

B

C

D

Solution

Given cubic polynomial $$(x-2)(x-4)(x-6)$$ has zeroes as $$x$$-coordinates of $$(2, 0) (4, 0)$$ and $$(6, 0)$$.
So, the third graph has these zeroes.
Option C is correct.
Question 6
1 / -0

A certain polynomial, P, has a degree of $$2$$. Polynomial P has zeros of $$2$$ and $$-3$$, and $$a>0$$ when the function of polynomial P is written in the form of $$y=ax^2+bx+c$$. Given this information, which of the following could be the graph of polynomial $$P?$$

A

B

C

D

Solution

Since the quadratic polynomial has zeroes as $$2$$ $$\&$$ $$-3$$ $$\&$$ the form of quadratic polynomial is $$ax^2+bx+c$$ $$\&$$ function of polynomial, $$y=ax^2+bx+c$$
The means, at $$x=2$$ $$\&$$ $$x=-3,$$ $$y$$ will be zero i.e, the function will take value as zero.
So, according to options given, option (c) and option (d) can be eliminated.
Now, to find whether the parabola is open upwards or downwards, we can make use of the additional information.
Since, i.e, $$a>0,$$ the coefficient of $$x^2$$ is positive.
At large values of $$x$$ on both sides of $$0$$(zero), the value of $$ax^2+bx+c$$ will be dominated by $$ax^2,$$ since the term has $$x^2$$ in it. So, for large values of $$x$$ if $$a>0,$$ then the function should always increase.
Hence, the parabola must open upwards.
Hence, the answer is option B.
Question 7
1 / -0

Which of the following is a polynomial with only one zero?

A
$$p(x)=2{ x }^{ 2 }-3x+4$$

B
$$p(x)={ x }^{ 2 }-2x+1$$

C
$$p(x)=2x+3$$

D
$$p(x)=5$$

Solution

Linear polynomial has only one zero.

Out of given options, $$p(x)=2x+3$$ is only linear polynomial.

Therefore, Option C is correct.
Question 8
1 / -0

Find the degree of the given algebraic expression:
$$ 3x-15$$

A
$$1$$

B
$$3$$

C
$$2$$

D
$$-15$$

Solution

The given algebraic expression $$3x-15$$ has two terms. The first one is $$3x$$ and the second is $$-15$$.

The exponent of the first term is $$1$$ because $$3x = 3x^1$$

The exponent of the second term is $$0$$ because $$15 = 15x^0$$

Since the highest exponent is $$1$$, therefore, the degree of $$3x-15$$ is $$1$$.

Hence, the degree of the algebraic expression $$3x-15$$ is $$1$$.
Question 9
1 / -0

Find $$X$$. If $$16.7+12.38-X=10.09$$.

A
$$17.89$$

B
$$18.99$$

C
$$16.98$$

D
$$20.09$$

Solution

Given equation is $$16.7+12.38-X=10.09$$
Shifting $$X$$ to the RHS, we get the following equation.
$$\Rightarrow 16.7+12.38-10.09=X$$
$$\Rightarrow X=18.99$$
Hence, Option B is correct.
Question 10
1 / -0

From the graph given below, $$y=P(x)$$ has .................. zeros.

A
$$1$$

B
$$5$$

C
$$3$$

D
$$4$$

Solution

Zeros of $$y=P(x)$$ can be obtained by equating $$y=P(x)=0$$
So, we need to find the points where $$y=0$$.
The point of intersection with $$x-$$axis gives the number of zeros.
In the above graph, $$P(x)$$ intersects the $$x-$$axis at four different points.
Hence, $$P(x)$$ has $$4$$ zeros.