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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Factorisation Test - 20

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Factorisation Test - 20

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Question 1

(8x2−8y2)÷8=x2−y2\displaystyle \left( 8{ x }^{ 2 }-8{ y }^{ 2 } \right) \div 8={ x }^{ 2 }-{ y }^{ 2 }(8x2−8y2)÷8=x2−y2

(8x2y2−16xy)÷8xy=(xy−2)\displaystyle \left( 8{ x }^{ 2 }{ y }^{ 2 }-16xy \right) \div 8xy=\left( xy-2 \right) (8x2y2−16xy)÷8xy=(xy−2)

(a2bc+ab2c+abc2)÷abc=(a+b+c)\displaystyle \left( { a }^{ 2 }bc+a{ b }^{ 2 }c+ab{ c }^{ 2 } \right) \div abc=\left( a+b+c \right) (a2bc+ab2c+abc2)÷abc=(a+b+c)

(a2bc+ab2c+abc2+abc)÷abc=(a+b+c)\displaystyle \left( { a }^{ 2 }bc+a{ b }^{ 2 }c+ab{ c }^{ 2 }+abc \right) \div abc=\left( a+b+c \right) (a2bc+ab2c+abc2+abc)÷abc=(a+b+c)

Solution

Question 2

(7x2−7)÷7=x2−7\displaystyle \left( 7{ x }^{ 2 }-7 \right) \div 7={ x }^{ 2 }-7(7x2−7)÷7=x2−7

(5x2+10)÷5=x2+10\displaystyle \left( 5{ x }^{ 2 }+10 \right) \div 5={ x }^{ 2 }+10(5x2+10)÷5=x2+10

(4x2+12)÷2=x2+6\displaystyle \left( 4{ x }^{ 2 }+12 \right) \div 2={ x }^{ 2 }+6(4x2+12)÷2=x2+6

(6x2+12)÷6=x2+2\displaystyle \left( 6{ x }^{ 2 }+12 \right) \div 6={ x }^{ 2 }+2(6x2+12)÷6=x2+2

Solution

Question 3

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

2xy+4\displaystyle 2xy+42xy+4

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

2x(y−5)\displaystyle 2x(y-5)2x(y−5)

Question 4

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

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

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

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

Solution

Question 5

2x\displaystyle 2x2x

2y\displaystyle 2y2y

2z\displaystyle 2z2z

xyz\displaystyle xyzxyz

Solution

Question 6

6(p−4q)\displaystyle 6(p-4q)6(p−4q)

6(p−q)\displaystyle 6(p-q)6(p−q)

6(1−4q)\displaystyle 6(1-4q)6(1−4q)

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

Solution

Question 7

35(x2+2x+4)\displaystyle 35\left( { x }^{ 2 }+2x+4 \right) 35(x2+2x+4)

5x2+2x+4\displaystyle 5{ x }^{ 2 }+2x+45x2+2x+4

x2+2x+4\displaystyle { x }^{ 2 }+2x+4x2+2x+4

5(x2+2x+4)\displaystyle 5\left( { x }^{ 2 }+2x+4 \right) 5(x2+2x+4)

Solution

Question 8

23\dfrac 2332​

10ab3\dfrac {10ab}{3}310ab​

10ab\displaystyle 10ab10ab

15ab2\dfrac {15ab}{2}215ab​

Solution

Question 9

6xy\displaystyle 6xy6xy

4xy\displaystyle 4xy4xy

x−4\displaystyle x-4x−4

xy(x−4)\displaystyle xy(x-4)xy(x−4)

Solution

Question 10

14\dfrac 1441​

−12-\dfrac 12−21​

−14-\dfrac 14−41​

12\dfrac 1221​

Solution

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

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

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

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

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

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

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

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

$\displaystyle 2x(y+4)$

$\displaystyle 2xy+4$

$\displaystyle 2y(y+4)$

$\displaystyle 2x(y-5)$

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

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

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

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

$\displaystyle 2x$

$\displaystyle 2y$

$\displaystyle 2z$

$\displaystyle xyz$

$\displaystyle 6(p-4q)$

$\displaystyle 6(p-q)$

$\displaystyle 6(1-4q)$

$\displaystyle 3(2-12q)$

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

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

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

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

$\dfrac 23$

$\dfrac {10ab}{3}$

$\displaystyle 10ab$

$\dfrac {15ab}{2}$

$\displaystyle 6xy$

$\displaystyle 4xy$

$\displaystyle x-4$

$\displaystyle xy(x-4)$

$\dfrac 14$

$-\dfrac 12$

$-\dfrac 14$

$\dfrac 12$