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11 April, 17:10

Factor completely 21x3 + 35x2 + 9x + 15. (3x - 5) (7x2 - 3) (3x - 5) (7x2 + 3) (3x + 5) (7x2 - 3) (3x + 5) (7x2 + 3)

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  1. 11 April, 18:15
    0
    Cubic = (3x + 5) (7x^2 + 3)

    Step-by-step explanation:

    If you make up two groups of 2, you might be able to factor this using the distributive property. Let's try it.

    First group: 21x^3 + 35x^2 Take out 7x^2 as a common fact

    First group: 7x^2 (3x + 5)

    Second group: 9x + 15 The HCF is 3

    Second group: 3 (3x + 5)

    Now put your factors together in one long string.

    Cubic = group 1 + group 2

    Cubic = 7x^2 (3x + 5) + 3 (3x + 5)

    Note: If you had something like 7x^2 * y + 3 * y then you should see that the factors are y * (7x^2 + 3). So to solve the cubic, you should observe that (3x + 5) is a common factor.

    Let y = 3x + 5

    y (7x^2 + 3) Now substitute back for the y.

    Cubic = (3x + 5) (7x^2 + 3)
  2. 11 April, 20:03
    0
    (7x^2+3) (3x+5)

    Step-by-step explanation:

    21x^3 + 35x^2 + 9x + 15.

    We will do factor by grouping

    Rearranging the terms so I have the terms with a factor of 3 first and 5 last

    Factor a 3x from the first two terms and 5 from the last two terms

    21x^3 + 9x + 35x^2 + 15

    3x (7x^2 + 3) + 5 (7x^2+3)

    Now factor out a (7x^2+3)

    (7x^2+3) (3x+5)
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