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

Nov 24, 2015

$\left(2 b - 1\right) \left(4 b + 3\right)$

#### Explanation:

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 $\to \left\{1 , 8\right\} , \left\{2 , 4\right\}$

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

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$\textcolor{b l u e}{\text{Attempt} \textcolor{w h i t e}{.} 1}$

$\textcolor{b r o w n}{\left(2 b + 1\right) \left(4 b - 3\right) \textcolor{b l u e}{= 8 {b}^{2} - 6 b + 4 b - 3} = 8 {b}^{2} - 2 b - 3} \textcolor{g r e e n}{\text{ Fail}}$

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$\textcolor{b l u e}{\text{Attempt} \textcolor{w h i t e}{.} 2}$

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

$\textcolor{b r o w n}{\left(2 b - 1\right) \left(4 b + 3\right) \textcolor{b l u e}{= 8 {b}^{2} - 4 b + 6 b - 3}}$

$\textcolor{b r o w n}{= 8 {b}^{2} + 2 b - 3} \textcolor{g r e e n}{\text{ It works!}}$

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