Factorisation Test

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Factorisation Test - 15

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Weekly Quiz Competition

Question 1
1 / -0

The expression $$\displaystyle xy-xz$$ is equivalent to:

A
$$\displaystyle x\left( y-z \right) $$

B
$$y(z-x)$$

C
$$x(y+z)$$

D
$$z(-x+y)$$

Solution

$$xy - xz $$
$$= x(y - z)$$
A number besides a bracket means the number is multiplied with each term in the bracket.
Question 2
1 / -0

Simplify: $$\displaystyle \frac{12ab^3-6a^2b}{3ab}$$ (given that $$ab\neq 0$$)

A
$$6b^2 - a$$

B
$$2b^2 - 3a$$

C
$$4b^2-2a$$

D
$$3b^2 - 2a$$

Solution

$$(2b^2 - a)$$ or $$4b^2 -2a$$: First, factor out common terms in the numerator. Then, cancel terms in both the numerator and denominator:
$$\displaystyle \frac{6ab(2b^2 -a)}{3ab} = 2(2b^2 - a)$$ or $$4b^2-2a$$
Question 3
1 / -0

If $$30x^{3} + 45x^{2} - 10x$$ is divided by $$5x$$, find the resulting coefficient of $$x$$

A
$$6$$

B
$$9$$

C
$$25$$

D
$$40$$

Solution

The value of $$\dfrac{30x^{3}+45x^{2}-10x}{5x}$$
$$=$$ $$\dfrac{5x\left ( 6x^{2}+9x-10 \right )}{5x}$$
$$=$$ $$6x^{2}+9x-10$$
In resultant equation, coefficient of $$x$$ is $$9$$.
Question 4
1 / -0

The expression $$\displaystyle abc+xyc$$ is equivalent to:

A
$$\displaystyle c\left( ab+xy \right) $$

B
$$c(ab-xy)$$

C
$$x(ab+cy)$$

D
$$x(ab-cy)$$

Solution

$$abc + xyc$$
$$= c(ab + xy)$$ ..... $$c$$ is common to both the terms
A number besides a bracket means the number is multiplied with each term in the bracket.
Question 5
1 / -0

Which of the following statements is true?

A
$$\displaystyle 5x+3=5(x+3)$$

B
$$\displaystyle 5x+3{ x }^{ 2 }=x\left( 5+3x \right) $$

C
$$\displaystyle 5x+3{ x }^{ 2 }=8x\left( x+{ x }^{ 2 } \right) $$

D
$$\displaystyle 5x+3=5\left( 1+3x \right) $$

Solution

$$\displaystyle 5x+3{ x }^{ 2 }=\left( 5\times x \right) +\left( 3\times x\times x \right) $$
$$ =x\left( 5+3x \right) $$

$$\displaystyle \therefore$$ The statement $$5x+3{ x }^{ 2 }=x\left( 5+3x \right) $$ is true.
Question 6
1 / -0

$$\left( 49{ x }^{ 2 }yz+35p \right) \div 7=$$

A
$$7{ x }^{ 2 }yz-35p$$

B
$$7{ x }^{ 2 }yz+35p$$

C
$$7{ x }^{ 2 }yz-5p$$

D
$$7{ x }^{ 2 }yz+5p$$

Solution

$$(49x^{2}yz+35p)\div 7$$
$$=(49x^{2}yz+35p)\times \frac{1}{7}$$
$$=\dfrac{49}{7}x^{2}yz+\dfrac{35}{7}p$$
$$=7x^{2}yz+5p$$
Question 7
1 / -0

Factorise completely by removing a monomial factor.
$$3{y^2} - 7y$$

A
7y(3y-1)

B
y(3y-7)

C
3y(y-7)

D
7(3y-1)

Solution

$$3y^2-7y$$
$$=y(3y-7)$$.
Question 8
1 / -0

One of the zeros of the polynomial $$2x^2+7x-4$$ is:

A
$$2$$

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

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

D
$$-2$$

Solution

To find the zero of $$2x^2+7x-4$$ is same as solving the equation $$2x^2+7x-4=0$$
$$\Rightarrow 2x^2+8x-x-4=0$$
$$\Rightarrow 2x(x+4)-1(x+4)=0$$
$$\Rightarrow (2x-1)(x+4)=0$$
$$\Rightarrow (2x-1)=0, (x+4)=0$$
$$\Rightarrow x=\dfrac12,-4$$
Therefore, option $$B$$ is correct.
Question 9
1 / -0

What is the possible expression for the dimension of the cuboid whose volume is given below?
$$Volume: 3x^2-12x$$

A
$$4\times x \times (x-4)$$

B
$$3\times x \times (x-4)$$

C
$$4\times x \times (x+4)$$

D
$$3\times x \times (x+4)$$

Solution

Given, $$Volume= 3x^2-12x$$
$$=3x(x-4)=3\times x\times(x-4)$$

$$\therefore$$Possible dimensions are 3 units, x units and (x-4) units
Question 10
1 / -0

Simplify :
$$10 a^2-15 b^2 + 20 c^2$$

A
$$5(2a^2 - 3b^2 + 4c^2)$$

B
$$5(2a -3b^2 + 4c^2)$$

C
$$5(10a^2 - 3b^2 + 4c^2)$$

D
$$5(2a^2 + 3b^2 + 4c^2)$$

Solution

$$10a^{ 2 }-15b^{ 2 }+20c^{ 2 }$$
The common factor $$=5$$
Thus, $$10a^{ 2 }-15b^{ 2 }+20c^{ 2 }$$
$$=5(2a^{ 2 }-3b^{ 2 }+4c^{ 2 })$$

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Re-Attempt Weekly Quiz Competition

Factorisation Test - 15

Result

Factorisation Test - 15

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SHARING IS CARING

Question 1

$$\displaystyle x\left( y-z \right) $$

$$y(z-x)$$

$$x(y+z)$$

$$z(-x+y)$$

Solution

Question 2

$$6b^2 - a$$

$$2b^2 - 3a$$

$$4b^2-2a$$

$$3b^2 - 2a$$

Solution

Question 3

$$6$$

$$9$$

$$25$$

$$40$$

Solution

Question 4

$$\displaystyle c\left( ab+xy \right) $$

$$c(ab-xy)$$

$$x(ab+cy)$$

$$x(ab-cy)$$

Solution

Question 5

$$\displaystyle 5x+3=5(x+3)$$

$$\displaystyle 5x+3{ x }^{ 2 }=x\left( 5+3x \right) $$

$$\displaystyle 5x+3{ x }^{ 2 }=8x\left( x+{ x }^{ 2 } \right) $$

$$\displaystyle 5x+3=5\left( 1+3x \right) $$

Solution

Question 6

$$7{ x }^{ 2 }yz-35p$$

$$7{ x }^{ 2 }yz+35p$$

$$7{ x }^{ 2 }yz-5p$$

$$7{ x }^{ 2 }yz+5p$$

Solution

Question 7

7y(3y-1)

y(3y-7)

3y(y-7)

7(3y-1)

Solution

Question 8

$$2$$

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

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

$$-2$$

Solution

Question 9

$$4\times x \times (x-4)$$

$$3\times x \times (x-4)$$

$$4\times x \times (x+4)$$

$$3\times x \times (x+4)$$

Solution

Question 10

$$5(2a^2 - 3b^2 + 4c^2)$$

$$5(2a -3b^2 + 4c^2)$$

$$5(10a^2 - 3b^2 + 4c^2)$$

$$5(2a^2 + 3b^2 + 4c^2)$$

Solution

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