How do you factor the trinomial $8b^2+2b-3$ ?

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Answer 1 · 2015-11-24T14:24:25

$(2b-1)(4b+3)$

Only factors of 3 are {1,3} so this is fixed giving the format of:

$(?+1)(?+3)$ .......ignoring the signs

But we have a negative 3 so the signs for the factors need to opposites. One positive and the other negative!

Consider the factors of 8 $-> {1,8}, {2,4}$

So can we end up with 2 in 2b using these?

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$color(blue)("Attempt" color(white)(.) 1)$

$color(brown)((2b+1)(4b-3)color(blue)( = 8b^2 -6b+4b-3) =8b^2-2b-3) color(green)(" Fail")$

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$color(blue)("Attempt" color(white)(.) 2)$

In attempt 1 the 2 in $2b$ is negative so we have the correct number but the sign is wrong. Try reversing the signs in the brackets.

$color(brown)((2b-1)(4b+3)color(blue)( = 8b^2-4b+6b-3))$

$color(brown)(=8b^2+2b-3) color(green)(" It works!")$

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How do you factor the trinomial 8b^2+2b-38b^2+2b-3?