Factorisation Test

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Factorisation Test - 20

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following is incorrect?

A
$$\displaystyle \left( 8{ x }^{ 2 }-8{ y }^{ 2 } \right) \div 8={ x }^{ 2 }-{ y }^{ 2 }$$

B
$$\displaystyle \left( 8{ x }^{ 2 }{ y }^{ 2 }-16xy \right) \div 8xy=\left( xy-2 \right) $$

C
$$\displaystyle \left( { a }^{ 2 }bc+a{ b }^{ 2 }c+ab{ c }^{ 2 } \right) \div abc=\left( a+b+c \right) $$

D
$$\displaystyle \left( { a }^{ 2 }bc+a{ b }^{ 2 }c+ab{ c }^{ 2 }+abc \right) \div abc=\left( a+b+c \right) $$

Solution

$$\displaystyle \left( { a }^{ 2 }bc+a{ b }^{ 2 }c+ab{ c }^{ 2 }+abc \right) \div abc=\frac { { a }^{ 2 }bc+a{ b }^{ 2 }c+ab{ c }^{ 2 }+abc }{ abc } $$

$$\displaystyle =\frac { abc\left( a+b+c+1 \right) }{ abc } =a+b+c+1$$

$$\displaystyle \therefore \left( { a }^{ 2 }bc+a{ b }^{ 2 }c+ab{ c }^{ 2 }+abc \right) \div abc$$ is an incorrect statement.
Question 2
1 / -0

Which of the following statements is correct?

A
$$\displaystyle \left( 7{ x }^{ 2 }-7 \right) \div 7={ x }^{ 2 }-7$$

B
$$\displaystyle \left( 5{ x }^{ 2 }+10 \right) \div 5={ x }^{ 2 }+10$$

C
$$\displaystyle \left( 4{ x }^{ 2 }+12 \right) \div 2={ x }^{ 2 }+6$$

D
$$\displaystyle \left( 6{ x }^{ 2 }+12 \right) \div 6={ x }^{ 2 }+2$$

Solution

$$\displaystyle \left( 6{ x }^{ 2 }+12 \right) \div 6=\frac { 6{ x }^{ 2 }+12 }{ 6 }$$

$$=\dfrac { 6\left( { x }^{ 2 }+2 \right) }{ 6 } ={ x }^{ 2 }+2$$

$$\displaystyle \therefore \left( 6{ x }^{ 2 }+12 \right) \div 6$$ is a correct statement.
Question 3
1 / -0

The value of $$\displaystyle \frac { 28xy\left( y-5 \right) \left( y+4 \right) }{ 14y\left( y-5 \right)}$$ is

A
$$\displaystyle 2x(y+4)$$

B
$$\displaystyle 2xy+4$$

C
$$\displaystyle 2y(y+4)$$

D
$$\displaystyle 2x(y-5)$$
Question 4
1 / -0

Factorise : $$\displaystyle 40{ m }^{ 2 }n+50mn$$

A
$$\displaystyle 10mn(4mn+5n)$$

B
$$\displaystyle 10mn(2m+5)$$

C
$$\displaystyle 10mn(2m+10)$$

D
$$\displaystyle 10mn(4m+5)$$

Solution

$$\displaystyle 40{ m }^{ 2 }n+50mn=\left( 2\times 5\times 2\times 2\times m\times m\times n \right) +\left( 2\times 5\times 5\times m\times n \right) $$
$$\displaystyle =10mn\left( 4m+5 \right) $$
Question 5
1 / -0

Simplify: $$\displaystyle \frac { 20xyz\left( 4x+5y+6z \right) }{ xz\left( 40x+50y+60z \right)}$$

A
$$\displaystyle 2x$$

B
$$\displaystyle 2y$$

C
$$\displaystyle 2z$$

D
$$\displaystyle xyz$$

Solution

$$\displaystyle \frac { 20xyz\left( 4x+5y+6z \right) }{ xz\left( 40x+50y+60z \right) } =\frac { 20y\left( 4x+5y+6z \right) }{ 10\left( 4x+5y+6z \right) } =2y$$
Question 6
1 / -0

Factorisation of the expression $$\displaystyle 6p-24q$$ results in :

A
$$\displaystyle 6(p-4q)$$

B
$$\displaystyle 6(p-q)$$

C
$$\displaystyle 6(1-4q)$$

D
$$\displaystyle 3(2-12q)$$

Solution

$$\displaystyle 6p-24q=\left( 6\times p \right) -\left( 24\times q \right) $$
$$=\left( 6\times p \right) -\left( 6\times 4\times q \right) $$
$$=6\left( p-4q \right) $$
Question 7
1 / -0

Evaluate: $$\displaystyle \frac { 35\left( x-3 \right) \left( { x }^{ 2 }+2x+4 \right) }{ 7\left( x-3 \right) } $$

A
$$\displaystyle 35\left( { x }^{ 2 }+2x+4 \right) $$

B
$$\displaystyle 5{ x }^{ 2 }+2x+4$$

C
$$\displaystyle { x }^{ 2 }+2x+4$$

D
$$\displaystyle 5\left( { x }^{ 2 }+2x+4 \right) $$

Solution

$$\displaystyle \frac { 35\left( x-3 \right) \left( { x }^{ 2 }+2x+4 \right) }{ 7\left( x-3 \right) } =5( { x }^{ 2 }+2x+4) $$
Question 8
1 / -0

Divide $$\displaystyle 10{ a }^{ 2 }{ b }^{ 2 }\left( 5x-25 \right)$$ by $$15ab\left( x-5 \right) $$

A
$$\dfrac 23$$

B
$$\dfrac {10ab}{3}$$

C
$$\displaystyle 10ab$$

D
$$\dfrac {15ab}{2}$$

Solution

$$\displaystyle 10{ a }^{ 2 }{ b }^{ 2 }\left( 5x-25 \right) \div 15ab\left( x-5 \right) $$

$$\displaystyle =\frac { 10{ a }^{ 2 }{ b }^{ 2 }\left( 5x-25 \right) }{ 15ab\left( x-5 \right) } $$

$$\displaystyle =\frac { 2ab\times 5\left( x-5 \right) }{ 3\left( x-5 \right) } =\frac { 10ab }{ 3 } $$
Question 9
1 / -0

Divide $$\displaystyle 4{ x }^{ 2 }{ y }^{ 2 }\left( 6x-24 \right) \div 4xy\left( x-4 \right) $$

A
$$\displaystyle 6xy$$

B
$$\displaystyle 4xy$$

C
$$\displaystyle x-4$$

D
$$\displaystyle xy(x-4)$$

Solution

$$\displaystyle 4{ x }^{ 2 }{ y }^{ 2 }\left( 6x-24 \right) \div 4xy\left( x-4 \right) $$

$$\displaystyle =\frac { 4{ x }^{ 2 }{ y }^{ 2 }\left( 6x-24 \right) }{ 4xy\left( x-4 \right) } =\frac { xy\times 6\left( x-4 \right) }{ \left( x-4 \right) } $$

$$\displaystyle =6xy$$
Question 10
1 / -0

Simplify: $$\displaystyle \frac { 4-x }{ 4x-16 }$$

A
$$\dfrac 14$$

B
$$-\dfrac 12$$

C
$$-\dfrac 14$$

D
$$\dfrac 12$$

Solution

$$\displaystyle \frac { 4-x }{ 4x-16 } =\frac { -\left( x-4 \right) }{ 4\left( x-4 \right) } =\frac { -1 }{ 4 } $$

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Re-Attempt Weekly Quiz Competition

Factorisation Test - 20

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Factorisation Test - 20

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SHARING IS CARING

Question 1

$$\displaystyle \left( 8{ x }^{ 2 }-8{ y }^{ 2 } \right) \div 8={ x }^{ 2 }-{ y }^{ 2 }$$

$$\displaystyle \left( 8{ x }^{ 2 }{ y }^{ 2 }-16xy \right) \div 8xy=\left( xy-2 \right) $$

$$\displaystyle \left( { a }^{ 2 }bc+a{ b }^{ 2 }c+ab{ c }^{ 2 } \right) \div abc=\left( a+b+c \right) $$

$$\displaystyle \left( { a }^{ 2 }bc+a{ b }^{ 2 }c+ab{ c }^{ 2 }+abc \right) \div abc=\left( a+b+c \right) $$

Solution

Question 2

$$\displaystyle \left( 7{ x }^{ 2 }-7 \right) \div 7={ x }^{ 2 }-7$$

$$\displaystyle \left( 5{ x }^{ 2 }+10 \right) \div 5={ x }^{ 2 }+10$$

$$\displaystyle \left( 4{ x }^{ 2 }+12 \right) \div 2={ x }^{ 2 }+6$$

$$\displaystyle \left( 6{ x }^{ 2 }+12 \right) \div 6={ x }^{ 2 }+2$$

Solution

Question 3

$$\displaystyle 2x(y+4)$$

$$\displaystyle 2xy+4$$

$$\displaystyle 2y(y+4)$$

$$\displaystyle 2x(y-5)$$

Question 4

$$\displaystyle 10mn(4mn+5n)$$

$$\displaystyle 10mn(2m+5)$$

$$\displaystyle 10mn(2m+10)$$

$$\displaystyle 10mn(4m+5)$$

Solution

Question 5

$$\displaystyle 2x$$

$$\displaystyle 2y$$

$$\displaystyle 2z$$

$$\displaystyle xyz$$

Solution

Question 6

$$\displaystyle 6(p-4q)$$

$$\displaystyle 6(p-q)$$

$$\displaystyle 6(1-4q)$$

$$\displaystyle 3(2-12q)$$

Solution

Question 7

$$\displaystyle 35\left( { x }^{ 2 }+2x+4 \right) $$

$$\displaystyle 5{ x }^{ 2 }+2x+4$$

$$\displaystyle { x }^{ 2 }+2x+4$$

$$\displaystyle 5\left( { x }^{ 2 }+2x+4 \right) $$

Solution

Question 8

$$\dfrac 23$$

$$\dfrac {10ab}{3}$$

$$\displaystyle 10ab$$

$$\dfrac {15ab}{2}$$

Solution

Question 9

$$\displaystyle 6xy$$

$$\displaystyle 4xy$$

$$\displaystyle x-4$$

$$\displaystyle xy(x-4)$$

Solution

Question 10

$$\dfrac 14$$

$$-\dfrac 12$$

$$-\dfrac 14$$

$$\dfrac 12$$

Solution

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