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11 December, 12:32

Factorise each of the following algebraic expressions completely,

(a) 2k-6k2 + 4k3

(b) 2ax - 4ay + 3bx - 6by

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Answers (1)
  1. 11 December, 13:02
    0
    see explanation

    Step-by-step explanation:

    (a)

    Given

    2k - 6k² + 4k³ ← factor out 2k from each term

    = 2k (1 - 3k + 2k²)

    To factor the quadratic

    Consider the factors of the product of the constant term (1) and the coefficient of the k² term ( + 2) which sum to give the coefficient of the k - term ( - 3)

    The factors are - 1 and - 2

    Use these factors to split the k - term

    1 - k - 2k + 2k² (factor the first/second and third/fourth terms)

    1 (1 - k) - 2k (1 - k) ← factor out (1 - k) from each term

    = (1 - k) (1 - 2k)

    1 - 3k + 2k² = (1 - k) (1 - 2k) and

    2k - 6k² + 4k³ = 2k (1 - k) (1 - 2k)

    (b)

    Given

    2ax - 4ay + 3bx - 6by (factor the first/second and third/fourth terms)

    = 2a (x - 2y) + 3b (x - 2y) ← factor out (x - 2y) from each term

    = (x - 2y) (2a + 3b)
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