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Work out the value of 3p − 4t², when p = 6 and t = 5 - Junior Cycle Mathematics - Question 5 - 2023

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Question 5

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Work out the value of 3p − 4t², when p = 6 and t = 5. Multiply out and simplify (2x − 3)(4 − 5x + x²). Factorise fully 10de − df − 5ef + 2d².

Worked Solution & Example Answer:Work out the value of 3p − 4t², when p = 6 and t = 5 - Junior Cycle Mathematics - Question 5 - 2023

Step 1

Work out the value of 3p − 4t², when p = 6 and t = 5.

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Answer

To calculate the value of the expression, substitute the given values of p and t:

  1. Substitute p = 6 and t = 5 into the expression:

    3(6)4(5)23(6) - 4(5)^2

  2. Carry out the calculations:

    184(25)18 - 4(25)

  3. Now calculate:

    18100=8218 - 100 = -82

Thus, the answer is -82.

Step 2

Multiply out and simplify (2x − 3)(4 − 5x + x²).

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Answer

To multiply out and simplify the expression, use the distributive property:

  1. Start with the given expression:

    (2x3)(45x+x2)(2x - 3)(4 - 5x + x^2)

  2. Distribute (2x) to each term in the second parentheses:

    2x(4)2x(5x)+2x(x2)=8x10x2+2x32x(4) - 2x(5x) + 2x(x^2) = 8x - 10x^2 + 2x^3

  3. Now distribute (-3) to each term:

    3(4)+3(5x)3(x2)=12+15x3x2-3(4) + 3(5x) - 3(x^2) = -12 + 15x - 3x^2

  4. Combine all terms:

    2x310x2+15x122x^3 - 10x^2 + 15x - 12

The simplified expression is:

2x310x2+15x122x^3 - 10x^2 + 15x - 12

Step 3

Factorise fully 10de − df − 5ef + 2d².

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Answer

To factorise the expression, identify and factor out the common factors:

  1. Start with the expression:

    10dedf5ef+2d210de - df - 5ef + 2d^2

  2. Group terms to find common factors:

    (10de+2d2)+(df5ef)(10de + 2d^2) + (-df - 5ef)

  3. Factor each group:

    2d(5e+d)f(d+5e)2d(5e + d) - f(d + 5e)

  4. Notice the common factor (d+5e)(d + 5e):

    (d+5e)(2df)\Rightarrow (d + 5e)(2d - f)

Thus, the fully factorised form is:

(d+5e)(2df)(d + 5e)(2d - f)

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