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Multiply out and simplify $(x + 9)(2x - 1)$ - Junior Cycle Mathematics - Question 8 - 2016

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Question 8

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Multiply out and simplify $(x + 9)(2x - 1)$. Factorise fully $3ax + ay + 3cx + cy$.

Worked Solution & Example Answer:Multiply out and simplify $(x + 9)(2x - 1)$ - Junior Cycle Mathematics - Question 8 - 2016

Step 1

Multiply out and simplify $(x + 9)(2x - 1)$

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Answer

To multiply (x+9)(2x−1)(x + 9)(2x - 1), we use the distributive property (also known as the FOIL method for binomials):

  1. Multiply the first terms: x⋅2x=2x2x \cdot 2x = 2x^2

  2. Multiply the outer terms: x⋅(−1)=−xx \cdot (-1) = -x

  3. Multiply the inner terms: 9⋅2x=18x9 \cdot 2x = 18x

  4. Multiply the last terms: 9⋅(−1)=−99 \cdot (-1) = -9

Now, combine the like terms:

2x2+18x−x−9=2x2+17x−92x^2 + 18x - x - 9 = 2x^2 + 17x - 9

Hence, the simplified result is:

2x2+17x−92x^2 + 17x - 9

Step 2

Factorise fully $3ax + ay + 3cx + cy$

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Answer

To factorise the expression 3ax+ay+3cx+cy3ax + ay + 3cx + cy, we can group the terms:

  1. Group the first two and the last two terms: (3ax+ay)+(3cx+cy)(3ax + ay) + (3cx + cy)

  2. Factor out the common factors from each group: a(3x+y)+c(3x+y)a(3x + y) + c(3x + y)

  3. Now, we can factor out (3x+y)(3x + y): (3x+y)(a+c)(3x + y)(a + c)

Thus, the fully factorised form is:

(3x+y)(a+c)(3x + y)(a + c)

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