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Algebraic Expressions and Identities Test - 36

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Algebraic Expressions and Identities Test - 36
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  • Question 1
    1 / -0
    What is the product of $$(x-2x)^2$$?
    Solution
    $$(x-2x)^2 = (x-2x)(x-2x)$$
    $$=$$ $$x.x-x.2x-2x.x+2x.2x$$
    $$=$$ $$x^2-2x^2-2x^2+4x^2$$
    $$=x^2$$
  • Question 2
    1 / -0
    Evaluate: $$-3z^2(-4x^2+10z^4-4xz^5)$$
    Solution
    $$-3z^2(-4x^2+10z^4-4xz^5)$$
    Multiply each term of the parenthesis by the monomial keeping the addition or subtraction sign same.
    $$=$$ $$-3z^2.(-4x^2)+-3z^2.10z^4-(-3z^2).4xz^5$$
    $$=$$ $$12x^2z^2-30z^6+12xz^7$$
  • Question 3
    1 / -0
    Find the value of $$(2x+5)(2x-5)$$.
    Solution
    We know that binomial identities for $$(x+y)(x-y)=x^2-y^2$$
    $$\therefore (2x+5)(2x-5) = (2x)^2-5^2$$
    $$=$$ $$4x^2-25$$
  • Question 4
    1 / -0
    Multiply: $$6x^5(-3x^4-2x^3+x^2)$$
    Solution
    $$6x^5(-3x^4-2x^3+x^2)$$
    Multiply each term of the parenthesis by the monomial keeping the addition or subtraction sign same.
    $$=$$ $$6x^5.-3x^4-6x^5.2x^3+6x^5.x^2$$
    $$=$$ $$-18x^9-12x^8+6x^7$$
  • Question 5
    1 / -0
    Simplify the following expression: $$(s-r)^2-(s+r)^2$$
    Solution
    We know that binomial identities for $$(x+y)^2=x^2+2xy+y^2$$
    and $$(x-y)^2=x^2-2xy+y^2$$
    So, $$(s-r)^2-(s+r)^2$$ $$=$$ $$s^2-2sr+r^2-s^2-2sr-r^2$$
    $$=$$ $$-4sr$$
  • Question 6
    1 / -0
    The value of $$-3x(5k^2+y)$$ is
    Solution
    We need to find value of $$-3x(5k^2+y)$$
    Multiply $$-3x$$ to the inside bracket, we get
    =$$-15k^2x-3xy$$
  • Question 7
    1 / -0
    Multiply: $$4x^2z(3x^2y^2-4xz)$$
    Solution
    $$4x^2z(3x^2y^2-4xz)$$
    Multiply each term of the parenthesis by the monomial keeping the addition or subtraction sign same.
    = $$4x^2z.3x^2y^2-4x^2z.4xz$$
    = $$12x^4y^2z-16x^3z^2$$
  • Question 8
    1 / -0
    Simplify: $$7a^3(-4a^4+6a^5)$$
    Solution
    $$7a^3(-4a^4+6a^5)$$
    Multiply each term of the parenthesis by the monomial keeping the addition or subtraction sign same.
    $$=$$ $$7a^3.-4a^4+7a^3.6a^5$$
    $$=$$ $$-28a^7+42a^8$$
  • Question 9
    1 / -0
    Reduce the following expression using binomial identities: $$(3x+5)(3x+10)$$.
    Solution
    We know that binomial identities for $$(x+y)(x+z)=x^2+(y+z)x+yz$$
    So, $$x = 3x, y = 5, z = 10$$
    $$=$$ $$(3x)^2+(5+10)3x+5\times 10$$
    $$=$$ $$9x^2+45x+50$$
  • Question 10
    1 / -0
    If $$x\Box y=(x+y)^2-(x-y)^2$$. Then $$\sqrt 4\Box \sqrt 5=$$.
    Solution

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