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Factorisation Test - 13

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Factorisation Test - 13
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  • Question 1
    1 / -0
    Factorise $$-4a^2 + 4ab -4ca$$
    Solution
    We have, $$4a^2= 2 \times  2 \times a \times a,$$
    $$4ab= 2 \times 2 \times a \times  b $$
    and,$$ 4ca=2 \times 2 \times c \times  a$$
    The three terms have 2, 2 and a as common facotrs
    $$-4a^2 + 4ab -4ca = -(2 \times 2 \times a \times a ) + (2 \times 2 \times a \times b ) - ( 2 \times 2 \times c \times a )$$
    $$=2 \times 2 \times a \times ( - a + b -c)$$
    $$=4a ( -a + b -c)$$
    $$=-4a(a-b+c)$$
  • Question 2
    1 / -0
    Evaluate $$(x^3 + 2x^2 + 3x) \div 2x$$
    Solution
      $$(x^3+2x^2+3x)\div 2x$$
    $$=(x\times x\times x)+(2\times x\times x)+(3\times x)$$
    $$=\frac{x(x^2+2x+3)}{2x}$$
    $$=\frac{1}{2}(x^2+2x+3)$$
  • Question 3
    1 / -0
    Factorise the expression: $$7p^2 + 21q^2$$
    Solution
       $$7p^2+21q^2$$
    $$=7\times p\times p + 7\times 3\times q\times q$$
    $$=7(p^2+3q^2)$$
  • Question 4
    1 / -0
    Factorise the expression: $$ax^2+ bx$$
    Solution
    $$ax^2+bx $$
    $$= x(ax+b)$$
  • Question 5
    1 / -0
    Factorise the expression: $$2x^3 + 2xy^2 + 2xz^2$$
    Solution
       $$2x^3+2xy^2+2xz^2$$
    $$=2\times x\times x\times x + 2\times x\times y\times y+2\times x\times z\times z$$
    $$=2x(x^2+y^2+z^2)$$
  • Question 6
    1 / -0
    Evaluate $$(10x - 25) \div 5$$
    Solution
    $$(10 x - 25) \div 5 $$

         $$= \displaystyle \frac{10 x - 25}{5}$$

         $$ = \dfrac{5 \times (2x - 5)}{5} $$

         $$= 2x - 5$$
  • Question 7
    1 / -0
    Divide the given polynomial by the given monomial. $$(3y^8 -4y^6+ 5y^4) \div y^4$$
    Solution
      $$(3y^84y^6+5y^4)\div y^4$$
    $$3y^8=3\times y\times y\times y\times y\times y\times y\times y\times y$$
    $$4y^6=4\times y\times y\times y\times y\times y\times y$$
    $$5y^4=5\times y\times y\times y\times y$$
    $$=\frac{(3\times y\times y\times y\times y\times y\times y\times y\times y)-(4\times y\times y\times y\times y\times y\times y)+(5\times y\times y\times y\times y)}{y^4}$$
    $$=\frac { y^ 4[(3\times y\times y\times y\times y)-(4\times y\times y)+(5) ]}{ y^{ 4 } } $$
    $$=3y^4-4y^2+5$$
  • Question 8
    1 / -0
    Factorise: $$\displaystyle 5x-115$$
    Solution
    $$\displaystyle 5x-115=\left( 5\times x \right) -\left( 5\times 23 \right) $$
                    $$ =5(x-23)$$
  • Question 9
    1 / -0
    Factorize: $$\displaystyle 2x+6$$
    Solution
    $$2x+6 = (2\times x) + (2 \times 3)$$
                 $$=2 (x + 3)$$
  • Question 10
    1 / -0
    Factorise:
    $$16a^2-24ab$$
    Solution
    $$16a^2-24ab$$
    Taking 8a as a common,
    $$=8a(2a-3b)$$
    Answer $$8a(2a-3b)$$
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