Now, lastly, step number four tells us that we should be able to see an obvious common factor, which we have. From the last two terms, we can factor out a 2, which would give us \(2(2x – 3)\). 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).
#Factor quadratic equation how to
Let’s look at how to factor the equation \(2x^2 + x – 6\) using these four steps. The two terms, at this point, should now have an obvious common factor.Group together the first two terms and the last two terms, then factor them individually.Well, let’s look at an example of one of these equations that may be a little trickier to factor. So, it may be we don’t just know the factors off the top of our head, and guessing might not be as quick as we would hope. However, our numbers aren’t always so straightforward. In our last example, it was relatively simple to find a common factor for the two numbers 8 and 16. When we look at the graph for \(8x^2+ 16x = 0\), we can see that it’s zero at \(x = 0\) and \(x = -2\).Īlright, so that is kind of a side note to answer the question “why does factoring matter?”īut now, let’s look back at how to actually factor. To do that, we would set our factors equal to zero and solve. Well, factoring the quadratic equation then sets us up to be able to find out where exactly our roots are, and our roots just mean where our graph is equal to zero. Well, if you recall a quadratic equation is always a parabola (or U-shaped graph). Maybe you’re asking, why on earth do I even need to factor? Why can’t I just leave it as it is? \(x^2\) and \(x\) share a common factor of \(x\). But, we still have something that can be factored out. Now, we have \(8(x^2 + 2x)\) being multiplied by everything on the inside. So, we can go ahead and factor out that 8. Well, 8 and 16 share a common factor of 8. What are the common factors of \(8x^2 + 16x = 0\)? Say we have the equation \(8x^2 + 16x = 0\) The easiest way to do this is to find the common factor. Now, expanding can be pretty easy we know exactly what to do to expand them when given our factors, but figuring out how to factor our expanded version can be a little harder. So, again we have our factors \((x + 2)(x + 6)\) on the left, and when you multiply that you get the expanded version: \((x^2+ 8x + 12)\). Once you multiply together you get \(x^2+ 8x + 12\). The actual quadratic equation is the expanded, or multiplied out version, of your two factors that are being multiplied.įor example, \((x + 2)\) and \((x + 6)\) are my factors that are being multiplied together. In order to factor a quadratic, you just need to find what you would multiply by in order to get the quadratic.
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