Polynomials Test

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Polynomials Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

Find the missing term in the following problem:
$$\left(\displaystyle \frac{3x}{4}\, -\, \displaystyle \frac{4y}{3} \right )^2\, =\, \displaystyle \frac{9x^2}{16}\, +\, ..........\, +\, \displaystyle \frac{16y^2}{9}$$.

A
$$2xy$$

B
$$- 2xy$$

C
$$12xy$$

D
$$- 12xy$$

Solution

Given, $$\left(\displaystyle \frac{3x}{4}\, -\, \displaystyle \frac{4y}{3} \right )^2\, $$.
We know, $$(a-b)^2=a^2-2ab+b^2$$.
Then,
$$\left(\displaystyle \frac{3x}{4}\, -\, \displaystyle \frac{4y}{3} \right )^2\, $$
$$=\left( \displaystyle \frac{3x}{4} \right )^2\, -\, 2\, \left(\displaystyle \frac{3x}{4} \right )\left(\displaystyle \frac{4y}{3} \right )\, +\, \left(\displaystyle \frac{4y}{3}\right)^2$$
$$=\, \displaystyle \frac{9x^2}{16}\, -\, 2xy\, +\, \displaystyle \frac{16y^2}{9}$$
$$=\, \displaystyle \frac{9x^2}{16}\, +\, (-\, 2xy)\, +\, \displaystyle \frac{16y^2}{9}$$.
Hence, the missing term is $$-2xy$$.
Therefore, option $$B$$ is correct.
Question 2
1 / -0

Simplify: $$(7m -8n)^2 + (7m + 8n)^2$$.

A
$$49m^2+164n^2$$

B
$$98m^2+128n^2$$

C
$$49m^2+64n^2$$

D
$$128m^2+64n^2$$

Solution

Given, $$(7m -8n)^2 + (7m + 8n)^2$$.

We know, $$(a-b)^2=a^2-2ab+b^2$$
and $$(a+b)^2=a^2+2ab+b^2$$.

Then,
$$(7m -8n)^2 + (7m + 8n)^2$$
$$ = ((7m)^2-2(7m)(8n) + (8n)^2)+((7m)^2 + 2(7m)(8n) + (8n)^2)$$
$$ = (49m^2-112mn + 64n^2)+(49m^2 + 112mn + 64n^2)$$
$$ = 49m^2-112mn + 64n^2+49m^2 + 112m + 64n^2$$
$$ = 98m^2 + 128n^2$$.

Therefore, option $$B$$ is correct.
Question 3
1 / -0

Simplify: $$3x(4x -5) + 3$$ and find its value for $$x =$$ $$\displaystyle \frac{1}{2}$$

A
$$-\dfrac{3}{2}$$

B
$$-\dfrac{1}{2}$$

C
$$-\dfrac{3}{7}$$

D
$$-\dfrac{3}{8}$$

Solution

We have, $$3x(4x -5)+ 3 =3x \times 4x-3x \times 5+3$$
$$ =12x^2-15x + 3$$
When $$x=\displaystyle \frac{1}{2}$$, then $$3x (4x-5) +3$$
$$=\displaystyle 3 \times \frac{1}{2} \left( 4 \times \frac{1}{2} - 5\right) + 3 $$
$$= \displaystyle \frac{3}{2} \times -3 + 3 = -\frac{9}{2} + 3 = -\frac{3}{2}$$
Question 4
1 / -0

The value of the polynomial $$x^2+5$$, at $$x=3$$ is

A
$$12$$

B
$$13$$

C
$$14$$

D
$$15$$

Solution

Let $$f(x)=x^2+5$$
Therefore, at $$x=3$$, $$f(x) = f(3)$$.
Hence, $$f(3) = 3^2+5 = 9+5 = 14$$
Question 5
1 / -0

The value of the polynomial $$7x^{4} + 6x^{2} + x$$ when $$x = -2$$ is

A
$$131$$

B
$$132$$

C
$$134$$

D
$$136$$

Solution

given, $$f(x) = 7x^4+6x^2+x$$

$$f(-2) = 7(-2)^4+6(-2)^2+(-2)$$

$$f(-2) = 7(16)+6(4)-2$$

$$f(-2) = 112+22 = 134$$
Question 6
1 / -0

The value of the polynomial $$4x^{2} + 3x - 7$$ at $$x=1$$ is:

A
$$1$$

B
$$2$$

C
$$0$$

D
None of the above

Solution

$$f(x) = 4x^2+3x-7$$

So at $$x=1$$,

$$f(1) = 4(1)^2+3(1)-7 $$

$$f(1) = 4+3-7 = 0$$
Question 7
1 / -0

If $$f(x) = -3x - 5$$, then the value of $$f(2)$$ is

A
$$-11$$

B
$$-1$$

C
$$1$$

D
$$11$$

E
$$5$$

Solution

Given, $$f(x)=-3x-5$$
Then $$F(2)=-3(2)-5$$
$$=-6-5$$
$$=-11$$
Question 8
1 / -0

If $$\displaystyle x = \frac{4}{3}$$ is a root of the polynomial $$f(x)= 6x^3-11x^2+ kx-20$$, then find the value of $$k$$.

A
$$17$$

B
$$19$$

C
$$22$$

D
$$23$$

Solution

$$f(x) = 6x^3-11x^2+ kx -20$$
$$\displaystyle f \left ( \frac{4}{3} \right ) = 6 \left ( \frac{4}{3} \right )^3 - 11 \left ( \frac{4}{3} \right )^2 + k \left ( \frac{4}{3} \right ) - 20 = 0$$
$$\Rightarrow \displaystyle 6 \left ( \frac{64}{27}\right) - 11 \left ( \frac{16}{9} \right ) + \frac{4k}{3} - 20 = 0$$
$$\Rightarrow 128-176+12k -180 = 0$$
$$\Rightarrow 12k+128 -356=0$$
$$ \Rightarrow 12k = 228$$
$$\Rightarrow k =19.$$
Question 9
1 / -0

Identify the degree of the given equation: $$\displaystyle { x }^{ 2 }+3x-5={ x }^{ 2 }+9x-23$$

A
Zero

B
One

C
Two

D
Three

Solution

$$x^2+3x-5=x^2+9x-23$$
We need to bring the given equation to it standard form.
$$\therefore -5+23=9x-3x$$ .....Transpose
$$\therefore 18 = 6x$$
$$\therefore x=3$$

Here, the highest power of variable $$x$$ is $$1$$.
Hence, the degree of the given equation is $$1$$.
Question 10
1 / -0

If $$g (x) = \displaystyle 3x^{2}-2x-5 $$, what is the value of $$g(-1)$$?

A
$$-4$$

B
$$-10$$

C
$$-6$$

D
$$0$$

Solution

Given, $$g(x)=x^{2}-2x-5$$
Then $$g(-1)=3(-1)^{2}-2(-1)-5$$
$$=3+2-5$$
$$=0$$

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Polynomials Test - 32

Result

Polynomials Test - 32

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SHARING IS CARING

Question 1

$$2xy$$

$$- 2xy$$

$$12xy$$

$$- 12xy$$

Solution

Question 2

$$49m^2+164n^2$$

$$98m^2+128n^2$$

$$49m^2+64n^2$$

$$128m^2+64n^2$$

Solution

Question 3

$$-\dfrac{3}{2}$$

$$-\dfrac{1}{2}$$

$$-\dfrac{3}{7}$$

$$-\dfrac{3}{8}$$

Solution

Question 4

$$12$$

$$13$$

$$14$$

$$15$$

Solution

Question 5

$$131$$

$$132$$

$$134$$

$$136$$

Solution

Question 6

$$1$$

$$2$$

$$0$$

None of the above

Solution

Question 7

$$-11$$

$$-1$$

$$1$$

$$11$$

$$5$$

Solution

Question 8

$$17$$

$$19$$

$$22$$

$$23$$

Solution

Question 9

Zero

One

Two

Three

Solution

Question 10

$$-4$$

$$-10$$

$$-6$$

$$0$$

Solution

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