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10r2 - 29r + 10 = (5r - 2)(2r - 5) when factored
Work:
First, the trinomial is factorable because ac = 100 = (-25)(-4) and b = -29 = (-25) + (-4).
I would like to multiply by 10/10 the trinomial:
10r2 - 29r + 10
= (10/10)(10r2 - 29r + 10)
= [(10r - 4)(10r - 25)]/(2)(5)
= (5r - 2)(2r - 5)
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∙ 12y ago
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