Factor the polynomial. x2−2x−15x^2-2x-15x2−2x−15

( x + 3 ) ( x − 5 ) \left(x+3\right)\left(x-5\right) ( x + 3 ) ( x − 5 )

Factor the polynomial. x 2 − 2 x − 15 x^2-2x-15 x 2 − 2 x − 15

Alright, everyone. Let's take a look at this practice problem here. Same thing. Can't use the greatest common factor, can't use grouping, and I also can't use a special formula. So how do I solve these? I'm going to use the AC method. So this polynomial is in standard form. My a is just equal to 1. My c is equal to negative 15 and my b term over here is negative 2. So the first thing I have to do is I have to build up my little table over here, and I'm looking for factors that multiply to the c term over here. Always take a look at the sign that's go that goes before because it's really important. So a times c, in other words, 1 times negative 15 is negative 15. So I'm looking for two numbers that multiply to negative 15. Let's go ahead and do that. Here I have let's just go start at 1. What do I have to multiply by 1 to get to negative 15? I actually multiply by negative 15. But then what happens is I wanna take the positive and negative factors of both of these, so I have to do negative one. And then what happens? Well, if I have negative one already, then I have to do positive 15 over here. So basically, what happens is whenever you have negative c terms, you actually have to flip one of the two numbers to be positive and negative. Right? So just 2 work, well, 2 doesn't go into 15. What about 3? 3 goes into 15 if I multiply it by negative 5, but then negative 3, I have to multiply by positive 5. Right? So you have to sort of flip the signs around. Okay. And then if I do 5, then I just get back to 3. So basically, these are all just the unique pairs. So let's add this up. I'm looking for 2 numbers that's from this list here that add to the b term, which in this case is negative 2. So b is negative 2. Okay. So 1 and negative 15 make negative 14, that doesn't work. Negative 115 make positive 14, that doesn't work. 3 and negative 5 make 2. So it looks like we're done here because negative 35 make positive 2, and that's not what we're looking for here. So we're done. My two numbers are gonna be positive 3 and negative 5. Those are my p and q values. Do you remember how this works out? This just works out to x+px+q over here, except one of these things is gonna be negative. So this whole thing actually just becomes x+3, and then x-five. So those are our two factors. Again, multiply this out. The 3 and the 5 multiply to negative 15. 3 and negative 5 also add to negative 2. So those are our 2 answers over here. Alright? It's all there is to it. Let me know if you have any questions. Thanks for watching.

Factor the polynomial. x 2 − 2 x − 15 x^2-2x-15 x 2 − 2 x − 15

( x + 3 ) ( x − 5 ) \left(x+3\right)\left(x-5\right) ( x + 3 ) ( x − 5 )

( x − 3 ) ( x − 5 ) \left(x-3\right)\left(x-5\right) ( x − 3 ) ( x − 5 )

( x − 3 ) ( x + 5 ) \left(x-3\right)\left(x+5\right) ( x − 3 ) ( x + 5 )

( x + 3 ) ( x + 5 ) \left(x+3\right)\left(x+5\right) ( x + 3 ) ( x + 5 )

0. Review of Algebra

Factoring Polynomials

College Algebra

0. Review of Algebra

Factoring Polynomials

Struggling with College Algebra?

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Multiple Choice

Factor the polynomial.
$x^2-2x-15$

$\left(x-3\right)\left(x-5\right)$

$\left(x-3\right)\left(x+5\right)$

$\left(x+3\right)\left(x-5\right)$

$\left(x+3\right)\left(x+5\right)$

Verified step by step guidance

Identify the polynomial to be factored: $x^2 - 2x - 15$.

Look for two numbers that multiply to the constant term (-15) and add to the linear coefficient (-2).

The numbers that satisfy these conditions are 3 and -5, since 3 * (-5) = -15 and 3 + (-5) = -2.

Rewrite the middle term using the numbers found: $x^2 + 3x - 5x - 15$.

Factor by grouping: Group the terms as $(x^2 + 3x) + (-5x - 15)$, then factor each group to get $x(x + 3) - 5(x + 3)$, and finally factor out the common binomial to get $(x + 3)(x - 5)$.

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