Factorisation Test

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Factorisation Test - 22

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Weekly Quiz Competition

Question 1
1 / -0

Factorise: $$\displaystyle 80{ p }^{ 2 }-72p$$

A
$$\displaystyle 8p(5p-9)$$

B
$$\displaystyle 16p(5p-4)$$

C
$$\displaystyle 8p(10p-8)$$

D
$$\displaystyle 8p(10p-9)$$

Solution

$$\displaystyle 80{ p }^{ 2 }-72p=\left( 8\times 10\times p\times p \right) -\left( 8\times 9\times p \right) $$

$$= 8p(10p-9)$$
Question 2
1 / -0

Factorisation of the expression $$\displaystyle 17{ p }^{ 2 }q-102p{ q }^{ 2 }$$ results in:

A
$$\displaystyle 17pq(p-8q)$$

B
$$\displaystyle 17pq(p-q)$$

C
$$\displaystyle 17pq(p-4q)$$

D
$$\displaystyle 17pq(p-6q)$$

Solution

$$\displaystyle 17{ p }^{ 2 }q-102p{ q }^{ 2 }$$

$$\displaystyle =\left( 17\times p\times p\times q \right) -\left( 17\times 6\times p\times q\times q \right) $$

$$\displaystyle =17pq(p-6q)$$
Question 3
1 / -0

The factorisation of $$ \left (21a^2+3a \right )$$ is

A
$$3a(7a+1)$$

B
$$7a(3a+1)$$

C
$$3a(7a+3a)$$

D
$$3a(a+7)$$

Solution

The factorisation of $$21a^2+3a$$ is $$3a(7a+1)$$
Taking common terms out, we get $$3a(7a+1)$$.
Question 4
1 / -0

Multiplying factors is an example of

A
polynomial

B
quadratic equation

C
division algorithm

D
factorisation

Solution

Multiplying factors is an example of factorisation.
Example: $$4x^2+2x$$ is a factor $$2x(2x+1)$$
By multiplying the factor we get $$2x(2x+1) = 4x^2+2x$$
Question 5
1 / -0

$$\left( 14{ x }^{ 2 }yz-28{ x }^{ 2 }{ y }^{ 2 }{ z }^{ 3 }+32{ y }^{ 2 }{ z }^{ 2 } \right) \div \left( -4xy \right) $$ is equal to

A
$$\dfrac { 7 }{ 2 } yz+7xy{ z }^{ 2 }+8xyz$$

B
$$-\dfrac { 7 }{ 2 } xz+7xy{ z }^{ 3 }-\dfrac { 8y{ z }^{ 2 } }{ x } $$

C
$$-\dfrac { 7 }{ 2 } xz-7xy{ z }^{ 3 }+\dfrac { 8y{ z }^{ 2 } }{ x } $$

D
$$\dfrac { 7 }{ 2 } xz-7xy{ z }^{ 2 }-\dfrac { 8y{ z }^{ 2 } }{ x } $$

Solution

$$(14x^{2}yz-28x^{2}y^{2}z^{3}+32y^{2}z^{2})\div (-4xy)$$

$$=\dfrac{14x^{2}yz}{-4xy}-\dfrac{28x^{2}y^{2}z^{3}}{-4xy}+\dfrac{32y^{2}z^{2}}{-4xy}$$

$$=-\dfrac{7}{2}xz+7xyz^{3}-\dfrac{8yz^{2}}{x}$$
Hence, B is correct.
Question 6
1 / -0

$$4a + 12b$$ is equal to

A
$$4a$$

B
$$12b$$

C
$$4(a+3b)$$

D
$$3a$$

Solution

Given: $$4a+12b$$
$$=4\cdot a+4\cdot 3\cdot b$$
$$=4\cdot a+4\cdot 3b$$
$$=4(a+3b)$$, (common factor $$4$$ is taken out)
Question 7
1 / -0

Directions For Questions

Divide the given polynomial by the given monomial.

...view full instructions

$$(4l^5 - 6l^4 + 8l^3) \div 2l^2$$

A
$$(2l^2 - 3l + 4)$$

B
$$l(2l^2 - 3l + 4)$$

C
$$l(2l^2 - 3l + 3)$$

D
None

Solution

We divide the given polynomial $$4l^5-6l^4+8l^3$$ by the monomial $$2l^2$$ as shown below:

$$\dfrac { 4l^{ 5 }-6l^{ 4 }+8l^{ 3 } }{ 2l^{ 2 } } \\ =\dfrac { 4l^{ 5 } }{ 2l^{ 2 } } -\dfrac { 6l^{ 4 } }{ 2l^{ 2 } } +\dfrac { 8l^{ 3 } }{ 2l^{ 2 } } \\ =\left( \dfrac { 2\times 2\times l^{ 5 } }{ 2l^{ 2 } } \right) -\left( \dfrac { 2\times 3\times l^{ 4 } }{ 2l^{ 2 } } \right) +\left( \dfrac { 2\times 2\times 2\times l^{ 3 } }{ 2l^{ 2 } } \right) \\ =\left( \dfrac { 2\times l^{ 5 }\times l^{ -2 } }{ 1 } \right) -\left( \dfrac { 3\times l^{ 4 }\times l^{ -2 } }{ 1 } \right) +\left( \dfrac { 2\times 2\times l^{ 3 }\times l^{ -2 } }{ 1 } \right)$$
$$=(2\times l^{ (5-2) })-(3\times l^{ (4-2) })+(4\times l^{ (3-2) })\quad \quad \quad \quad \quad \quad \left( \because \quad a^{ x }+a^{ y }=a^{ x+y } \right) \\ =2l^{ 3 }-3l^{ 2 }+4l\\ =l(2l^{ 2 }-3l+4)$$

Hence, $$\dfrac { 4l^{ 5 }-6l^{ 4 }+8l^{ 3 } }{ 2l^{ 2 } } =l(2l^{ 2 }-3l+4)$$.
Question 8
1 / -0

Directions For Questions

Divide the given polynomial by the given monomial.

...view full instructions

$$15(a^3 b^2 c^2 - a^2 b^3 c^2 + a^2b^2c^3) \div 3abc$$

A
$$abc(a - b + c)$$

B
$$5abc(a + b + c)$$

C
$$5abc(a - b + c)$$

D
None

Solution

We divide the given polynomial $$15(a^3b^2c^2-a^2b^3c^2+a^2b^2c^3)$$ by the monomial $$3abc$$ as shown below:

$$\dfrac { 15(a^{ 3 }b^{ 2 }c^{ 2 }-a^{ 2 }b^{ 3 }c^{ 2 }+a^{ 2 }b^{ 2 }c^{ 3 }) }{ 3abc } \\ =\dfrac { 15a^{ 3 }b^{ 2 }c^{ 2 }-15a^{ 2 }b^{ 3 }c^{ 2 }+15a^{ 2 }b^{ 2 }c^{ 3 } }{ 3abc } \\ =\dfrac { 15a^{ 3 }b^{ 2 }c^{ 2 } }{ 3abc } -\dfrac { 15a^{ 2 }b^{ 3 }c^{ 2 } }{ 3abc } +\dfrac { 15a^{ 2 }b^{ 2 }c^{ 3 } }{ 3abc } \\ =\left( \dfrac { 3\times 5\times a\times a\times a\times b\times b\times c\times c }{ 3abc } \right) -\left( \dfrac { 3\times 5\times a\times a\times b\times b\times b\times c\times c }{ 3abc } \right) +\left( \dfrac { 3\times 5\times a\times a\times b\times b\times c\times c\times c }{ 3abc } \right)$$
$$=5a^{ 2 }bc-5ab^{ 2 }c+5abc^{ 2 }\\ =5abc(a-b+c)$$

Hence, $$\dfrac { 15(a^{ 3 }b^{ 2 }c^{ 2 }-a^{ 2 }b^{ 3 }c^{ 2 }+a^{ 2 }b^{ 2 }c^{ 3 }) }{ 3abc } =5abc(a-b+c)$$.
Question 9
1 / -0

Directions For Questions

Divide the given polynomial by the given monomial.

...view full instructions

$$(3x^2 - 2x) \div x$$

A
$$3x$$

B
$$3x - 2$$

C
$$3x + 2$$

D
None

Solution

Let us first factorize the given polynomial $$3x^2-2x$$ by finding the HCF of all the terms as follows:

$$3x^{ 2 }=3\times x\times x\\ 2x=2\times x$$

Therefore, $$HCF=x$$

Now, we factor out the HCF from each term of the polynomial $$3x^2-2x$$ as shown below:

$$3x^{ 2 }-2x=x(3x-2)$$

Let us now divide the polynomial $$3x^2-2x$$ by the monomial $$x$$:

$$\dfrac { 3x^{ 2 }-2x }{ x } =\dfrac { x(3x-2) }{ x } =3x-2$$

Hence, $$(3x^{ 2 }-2x)\div x=3x-2$$.
Question 10
1 / -0

Directions For Questions

Divide the given polynomial by the given monomial.

...view full instructions

$$(3p^3 - 9p^2q - 6pq^2) \div (-3p)$$

A
$$(2q^2 + 3pq + p^2)$$

B
$$(2q^2 + pq - p^2)$$

C
$$(2q^2 + 3pq - p^2)$$

D
None

Solution

We divide the given polynomial $$3p^3-9p^2q-6pq^2$$ by the monomial $$-3p$$ as shown below:

$$\Rightarrow \dfrac { 3p^{ 3 }-9p^{ 2 }q-6pq^{ 2 } }{ -3p } \\ =\dfrac { 3p^{ 3 } }{ -3p } -\dfrac { 9p^{ 2 }q }{ -3p } -\dfrac { 6pq^{ 2 } }{ -3p } \\ =-\dfrac { 3p^{ 3 } }{ 3p } +\dfrac { 9p^{ 2 }q }{ 3p } +\dfrac { 6pq^{ 2 } }{ 3p } \\ =-\left( \dfrac { 3\times p\times p\times p }{ 3p } \right) -\left( \dfrac { 3\times 3\times p\times p\times q }{ 3p } \right) +\left( \dfrac { 2\times 3\times p\times q\times q }{ 3p } \right) \\ =-p^{ 2 }+3pq+2q^{ 2 }$$

Hence, $$\dfrac { 3p^{ 3 }-9p^{ 2 }q-6pq^{ 2 } }{ -3p } =2q^{ 2 }+3pq-p^{ 2 }$$.

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Factorisation Test - 22

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Factorisation Test - 22

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Question 1

$$\displaystyle 8p(5p-9)$$

$$\displaystyle 16p(5p-4)$$

$$\displaystyle 8p(10p-8)$$

$$\displaystyle 8p(10p-9)$$

Solution

Question 2

$$\displaystyle 17pq(p-8q)$$

$$\displaystyle 17pq(p-q)$$

$$\displaystyle 17pq(p-4q)$$

$$\displaystyle 17pq(p-6q)$$

Solution

Question 3

$$3a(7a+1)$$

$$7a(3a+1)$$

$$3a(7a+3a)$$

$$3a(a+7)$$

Solution

Question 4

polynomial

quadratic equation

division algorithm

factorisation

Solution

Question 5

$$\dfrac { 7 }{ 2 } yz+7xy{ z }^{ 2 }+8xyz$$

$$-\dfrac { 7 }{ 2 } xz+7xy{ z }^{ 3 }-\dfrac { 8y{ z }^{ 2 } }{ x } $$

$$-\dfrac { 7 }{ 2 } xz-7xy{ z }^{ 3 }+\dfrac { 8y{ z }^{ 2 } }{ x } $$

$$\dfrac { 7 }{ 2 } xz-7xy{ z }^{ 2 }-\dfrac { 8y{ z }^{ 2 } }{ x } $$

Solution

Question 6

$$4a$$

$$12b$$

$$4(a+3b)$$

$$3a$$

Solution

Question 7

$$(2l^2 - 3l + 4)$$

$$l(2l^2 - 3l + 4)$$

$$l(2l^2 - 3l + 3)$$

None

Solution

Question 8

$$abc(a - b + c)$$

$$5abc(a + b + c)$$

$$5abc(a - b + c)$$

None

Solution

Question 9

$$3x$$

$$3x - 2$$

$$3x + 2$$

None

Solution

Question 10

$$(2q^2 + 3pq + p^2)$$

$$(2q^2 + pq - p^2)$$

$$(2q^2 + 3pq - p^2)$$

None

Solution

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