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Factorise Completely a) y² - 16y b) 32x² - 25 c) 18x² - 100x

Mathematics

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Answers

3 parts
a

y(y−16)\mathbf{y(y - 16)}

b

(42x−5)(42x+5)\mathbf{(4\sqrt{2}x - 5)(4\sqrt{2}x + 5)}

c

2x(9x−50)\mathbf{2x(9x - 50)}

Step-by-step solution

7 steps

Here are the factorizations for the given expressions.

Part a· 2 steps

Factorise y2−16yy^2 - 16y

  1. Identify the common factor in both terms.

    Both terms, y2y^2 and 16y16y, have yy as a common factor.

  2. Factor out the common factor yy.

    y2−16y=y(y−16)y^2 - 16y = y(y - 16)

    The completely factorised expression is y(y−16)\boxed{\mathbf{y(y - 16)}}.

Part b

(42x−5)(42x+5)\mathbf{(4\sqrt{2}x - 5)(4\sqrt{2}x + 5)}

Factorise 32x2−2532x^2 - 25

  1. Recognize the form as a difference of squares, A2−B2=(A−B)(A+B)A^2 - B^2 = (A - B)(A + B).

    We can write 32x232x^2 as (32x)2(\sqrt{32}x)^2 and 25 as 525^2.

    32x2−25=(32x)2−5232x^2 - 25 = (\sqrt{32}x)^2 - 5^2
  2. Simplify 32\sqrt{32}.

    32=16×2=16×2=42\sqrt{32} = \sqrt{16 \times 2} = \sqrt{16} \times \sqrt{2} = 4\sqrt{2}
  3. Substitute the simplified radical back into the expression and apply the difference of squares formula.

    (42x)2−52=(42x−5)(42x+5)(4\sqrt{2}x)^2 - 5^2 = (4\sqrt{2}x - 5)(4\sqrt{2}x + 5)

    The completely factorised expression is (42x−5)(42x+5)\boxed{\mathbf{(4\sqrt{2}x - 5)(4\sqrt{2}x + 5)}}.

Part c

2x(9x−50)\mathbf{2x(9x - 50)}

Factorise 18x2−100x18x^2 - 100x

  1. Identify the greatest common factor (GCF) of the coefficients and the variables.

    The coefficients are 18 and 100. The GCF of 18 and 100 is 2.
    The variables are x2x^2 and xx. The GCF of x2x^2 and xx is xx.
    So, the overall GCF is 2x2x.

  2. Factor out the GCF $2x$ from both terms.

    18x2−100x=2x(9x−50)18x^2 - 100x = 2x(9x - 50)

    The completely factorised expression is 2x(9x−50)\boxed{\mathbf{2x(9x - 50)}}.

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