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Factorisation Test - 10

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Factorisation Test - 10
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  • Question 1
    1 / -0
    Factorise the following expression:
    $$7a^2 + 14a$$
    Solution
    Given expression is $$7a^2 + 14a$$
    Taking common factor $$7a$$ from both terms we get,
    $$7a(a+2)$$
    Hence, option A is correct.
  • Question 2
    1 / -0
    Factorise the following expressions:
    $$a x^2 y + b x y^2 + c x y z$$
    Solution
    $$a x^2 y + b x y^2 + c x y z$$
    In the above expression, the common factor is $$=xy$$
    $$\therefore$$ $$a x^2 y + b x y^2 + c x y z $$ = $$xy(ax+by+cz)$$
  • Question 3
    1 / -0
    Divide the given polynomial by the given monomial.
    $$(3y^8-  4y^6 + 5y^4) \div y^4$$
    Solution
        $$(3y^{ 8 }-4y^{ 6 }+5y^{ 4 })\div y^{ 4 }$$
    $$=\dfrac{(3y^{ 8 }-4y^{ 6 }+5y^{ 4 })}{y^{ 4 }}$$
    $$=\dfrac{y^4(3y^{ 4 }-4y^{ 2 }+5)}{y^{ 4 }}$$
    $$=(3y^{ 4 }-4y^{ 2 }+5)$$
  • Question 4
    1 / -0
    Factorise the expression.
    $$7p^2 + 21q^2$$
    Solution
    $$\begin{aligned}{}7{p^2} + 21{q^2}& = 7{p^2} + 7 \times 3{q^2}\\ &= 7\left( {{p^2} + 3{q^2}} \right)\end{aligned}$$
  • Question 5
    1 / -0
    Factorise the following expression:
    $$ 16 z + 20 z^3$$
    Solution
        $$16 z + 20 z^3$$
    Taking common factor 4z from both the terms we get,
    $$=4z(4+5z^2)$$
  • Question 6
    1 / -0
    Factorise the following expression:
    $$ 4 a^2 + 4 ab -4 ca$$
    Solution
    $$4a^{ 2 }+4ab-4ca$$
    The common factor $$=2\times 2\times a$$
    Thus, $$4a^{ 2 }+4ab-4ca=4a(a+b-c)$$
  • Question 7
    1 / -0
    Factorise the expression:
    $$ax^2 + bx$$
    Solution
    To factorise: $$ax^2+bx$$

    Now,
    $$ax^2+bx$$
    $$=ax\times x+b\times x$$
    $$=x(ax+b)$$

    Hence, the factorised form is $$x(ax+b)$$ .
  • Question 8
    1 / -0
    Factorize:
    $$5 x^2 y -15 xy^2$$
    Solution
    $$5{x  }^{  2}y-15x{y}^{2}$$
    Taking common factor $$5xy$$ for both the terms we get, 
    $$=5xy(x-3y)$$
  • Question 9
    1 / -0
    Factorise: $$ab-4ac$$
    Solution
    The common variable in the two terms is $$a$$. So, we can take $$a$$ common from both the terms as follows:
    $$ ab-4ac =a(b-4c)$$

    Hence, option C is correct.
  • Question 10
    1 / -0
    Factorise: $$4a + 12b$$
    Solution
    $$4a+12b$$
    $$= (2\times 2 \times a) + (2\times 2\times 3 \times b)$$
    $$= (2\times 2) (a + 3 \times b)$$
    $$= 4 (a+3b)$$
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