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Lesson by
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Mike McGarry
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Magoosh Expert

Magoosh Expert

Now that we have learned a few different factoring techniques. Sort of like a bag of tricks we have. We can explore methods for combining them to factor more challenging expressions. These combined techniques would be applicable only on the most difficult Quant problems on the test. So what I'm covering in this video is relatively advanced.

For example, suppose you had to factor something of this form. Well notice, first of all, we can use our old trick of factoring out a greatest common factor. And, of course, a greatest common factor, both coefficients are divisible by 6, and we can factor out an x as well.

So we factor out a 6x and we get x squared minus 25. Well, and of course now, that's a difference of two squares. So it factors to this. And this is the fully factored form. Suppose we had to factor something like this. So this isn't too hard.

This could appear in many places on the test. Something in this form. So notice first of all, all the coefficients are even, so we can factor out a 2. Once we factor out a 2, then we get an ordinary quadratic. Now it's just a matter of factoring that ordering quadratic.

We need a product of 24, a sum of negative 11. So those two terms are negative 3 and negative 8, and so this factors to 2 times x minus 8 times x minus 3. This one's a little bit trickier. And that's certainly a scary looking trinomial.

But notice that first of all, every coefficient is divisible by 7, so we can factor out a greatest common factor of 7 and also we can factor out an x squared, so we factor out 7x squared, then we get x squared minus 8x minus 9. So that now again, is an ordinary quadratic. We need a product of negative 9, and a sum of negative 8, so plus 1 and negative 9 would be the two numbers that would satisfy that.

And so this can be factored into 7x squared times x minus 9 times x plus 1. Another scary looking thing. Well, first of all, we can change, swap the order around so that we have a positive first, and a negative second. That makes things a little bit easier.

We can factor out, a factor of 3, and we can also factor out a single factor of x, that's the greatest common factor. That gets us down to 3x times the quantity 16y to the 4th minus x to the 8th. Well, those are both even powers so they're both squares and of course, 16 is a square, so 16y to the 4th would be the square of 4y squared and x to the 8th would be the square of x to the 4th.

So we can factor this into this. Now, notice that, that last parenthesis, 4y squared minus x to the 4th, that is another difference of squares. So we can factor that by the difference of squares again 4x squared is the square of 4y squared is the square of 2y, and x to the fourth is the square of x squared. And so all together, this factors to that.

Now this would be a very, very advanced problem that would only appear on the hardest part of the test. I'll also say the factoring these problems is about as hard as any factoring you'll need to do on the test. Keep in mind though, the test will rarely hand you something as straightforward as factor this.

Here's some ugly algebraic expression, factor it. That doesn't often happen. Factoring is unlikely to be the thrust of an entire problem but you will need to factor as a step in a larger complicated problem. So having these skills of factoring, even factoring these really hairy expressions, is something that can help you unlock those harder Quant problems.

We will see some examples of factoring use in equation solving in the section on Algebra Equations at the end of our section on Algebra Expressions. In this lesson we talked about combining the individual factoring techniques to factor more complicated expressions.

Show TranscriptFor example, suppose you had to factor something of this form. Well notice, first of all, we can use our old trick of factoring out a greatest common factor. And, of course, a greatest common factor, both coefficients are divisible by 6, and we can factor out an x as well.

So we factor out a 6x and we get x squared minus 25. Well, and of course now, that's a difference of two squares. So it factors to this. And this is the fully factored form. Suppose we had to factor something like this. So this isn't too hard.

This could appear in many places on the test. Something in this form. So notice first of all, all the coefficients are even, so we can factor out a 2. Once we factor out a 2, then we get an ordinary quadratic. Now it's just a matter of factoring that ordering quadratic.

We need a product of 24, a sum of negative 11. So those two terms are negative 3 and negative 8, and so this factors to 2 times x minus 8 times x minus 3. This one's a little bit trickier. And that's certainly a scary looking trinomial.

But notice that first of all, every coefficient is divisible by 7, so we can factor out a greatest common factor of 7 and also we can factor out an x squared, so we factor out 7x squared, then we get x squared minus 8x minus 9. So that now again, is an ordinary quadratic. We need a product of negative 9, and a sum of negative 8, so plus 1 and negative 9 would be the two numbers that would satisfy that.

And so this can be factored into 7x squared times x minus 9 times x plus 1. Another scary looking thing. Well, first of all, we can change, swap the order around so that we have a positive first, and a negative second. That makes things a little bit easier.

We can factor out, a factor of 3, and we can also factor out a single factor of x, that's the greatest common factor. That gets us down to 3x times the quantity 16y to the 4th minus x to the 8th. Well, those are both even powers so they're both squares and of course, 16 is a square, so 16y to the 4th would be the square of 4y squared and x to the 8th would be the square of x to the 4th.

So we can factor this into this. Now, notice that, that last parenthesis, 4y squared minus x to the 4th, that is another difference of squares. So we can factor that by the difference of squares again 4x squared is the square of 4y squared is the square of 2y, and x to the fourth is the square of x squared. And so all together, this factors to that.

Now this would be a very, very advanced problem that would only appear on the hardest part of the test. I'll also say the factoring these problems is about as hard as any factoring you'll need to do on the test. Keep in mind though, the test will rarely hand you something as straightforward as factor this.

Here's some ugly algebraic expression, factor it. That doesn't often happen. Factoring is unlikely to be the thrust of an entire problem but you will need to factor as a step in a larger complicated problem. So having these skills of factoring, even factoring these really hairy expressions, is something that can help you unlock those harder Quant problems.

We will see some examples of factoring use in equation solving in the section on Algebra Equations at the end of our section on Algebra Expressions. In this lesson we talked about combining the individual factoring techniques to factor more complicated expressions.