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4 June, 18:35

Factor the polynomial completetly 21b2-40b-21

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  1. 4 June, 20:28
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    First we note symmetry in the expression's coefficients.

    We also note that 7*3=21, and 7+3=10.

    From the rational roots theorem, we are tempted to try with 3 and 7 as coefficients of the factors.

    Try

    (7b+3) (3b+7) = 21b^2 + (49+9) b+21

    By switching the sign of 3b+7 to 3b-7, we get the signs right, to check:

    (7b+3) (3b-7) = 21b^2 + (9-49) b-21=21b^2-40b-21 ... right!

    So

    (7b+3) (3b-7) = 21b^2-40b-21
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