![]() From the first two terms, we can factor out an \(x\), which would give us \(x(2x – 3)\). So, we would group together \((2x^2 – 3x) + (4x-6)\), then factor. Step 3 tells us to group together the first two terms, and the last two terms, then factor them individually. Step two tells us to replace the middle of the equation with the two factors that we found. That makes our factors -3 and 4, which multiplies to give us -12 and add to give us positive 1. ![]() So, the only way that can happen is if we take the two factors 3 and 4, and throw a negative sign in front of 3. So, we need these factors to multiply to get -12, which each of them will if we through a negative sign in front of one of the factors, but we also need for it to add to get positive one. Remember, we are actually dealing with a -12, so we need to consider that when choosing which two factors work. So, obviously there is 1 and 12, then 2 and 6, and lastly 3 and 4. So, let’s go ahead and list all the possible factors of 12. So, first, we need to identify which two numbers multiply together to get \(a\times c\), which is -12, and add to get \(b\) (positive 1). Let’s look at how to factor the equation \(2x^2 + x – 6\) using these four steps.
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