Factor the polynomial by grouping−x2−5x+7x+35-x^2-5x+7x+35−x2−5x+7x+35

( − x + 7 ) ( x + 5 ) \left(-x+7\right)\left(x+5\right) ( − x + 7 ) ( x + 5 )

Factor the polynomial by grouping − x 2 − 5 x + 7 x + 35 -x^2-5x+7x+35 − x 2 − 5 x + 7 x + 35

Alright, everyone. Let's take a look at this practice problem. We're gonna factor this polynomial by grouping. So hopefully, you have your table of your different factoring methods hand handy because we're just gonna jump right in and stick to the steps. The first thing we're gonna do here is make sure that our polynomial is written in standard form, which it is. In this case, we have decreasing exponents. We have something to the second power and to the first power and then so on and so forth. That's the first thing. The next thing we're gonna do is we're gonna separate these these pairs of terms of these terms into 2 pairs. And usually, the first and last pairs are gonna be perfectly fine to do this. So if you take a look what happens here, we have 2 pairs. And in this first pair, we have to figure out what's common between them, and the second pair, we have to do the same. That's the third step. We're gonna factor in a one term greatest common factor from each one of the groups. So let's take a look at how this, how this breaks down. If I have a negative x squared, and remember, I like to have a parenthesis, and if I have a negative, just put a -one just so you don't forget about it. This is -one times x times x. And then over here, I have a 5 over here times x. Now the 5 can't break down any further, but I do notice that I have at least 1 power of x in each one of the terms. So that means that my greatest common factor in this first pair over here is just x. So I'm gonna pull it to the outside of the parenthesis, cancel that out over here, and what I'm left with is I'm just left with x, then I have negative x over here. That's what's left over here. And then I have minus 5 over here. So that's the first term or the first group. Now I'll just do the same thing for the second group, see what happens. So this expression over here breaks down to 7 times x. I can't break 7 down any further because it's a prime number. What about 35? We'll notice that 7 is actually a a multiple or sorry 35 is a multiple of 7. So we could do is we can say this is actually just when you break this down 7 times 5. And really clearly here, we can see that 7 is actually gonna be our greatest common factor out of the second pair. So we'll pull this to the outside of the parentheses over here, cancel it out, and what we're left with here is plus 7 that gets brought to the outside of the parentheses. What's left on the inside? I just have x plus 5. Okay. So if you notice here, I've done that 3rd step. I've factored out the, greatest common factors, and usually what we've seen is that we end up with the same exact term on the first and second pair, but that didn't happen here. Here we got a negative x minus 5, and here we got an x+5. So what gives? How do we actually do this? Well, it turns out that you can actually because both of these terms are negative, you can actually pull out a negative one to the outside of both of these terms. Basically, what happens is you could rewrite this as negative one times x, and then plus negative one times five. So now what happens is there's an extra negative one that you can pull out to the outside of the parenthesis. And what really this becomes is this becomes negative x, x plus 5. So that is how this expression reduces. And then you have plus 7x+5. So and I didn't do anything to the second pair over here. So now that you've seen that I've basically just pulled the minus sign to the outside of the expression, now what I have is I have the 2 pairs that have the same common factor now. Now I have the x plus fives that are the same. So now what you can do here is we can take this x+5 and move it to the outside of the parenthesis, and we'll see that our final answer becomes x+5, and then we have whatever's left on the inside. I have negative x+7 So negative x plus 7. Alright. And then you'll and if you foil this out and if you multiply these 2 polynomials out, you should get back to your original expression over here. Alright? So that's it for this one. Thanks for watching.

Factor the polynomial by grouping − x 2 − 5 x + 7 x + 35 -x^2-5x+7x+35 − x 2 − 5 x + 7 x + 35

( − x + 7 ) ( x + 5 ) \left(-x+7\right)\left(x+5\right) ( − x + 7 ) ( x + 5 )

( x + 7 ) ( x + 5 ) \left(x+7\right)\left(x+5\right) ( x + 7 ) ( x + 5 )

( x − 7 ) ( x + 5 ) \left(x-7\right)\left(x+5\right) ( x − 7 ) ( x + 5 )

( x + 7 ) ( − x + 5 ) \left(x+7\right)\left(-x+5\right) ( x + 7 ) ( − x + 5 )

0. Fundamental Concepts of Algebra

Factoring Polynomials

Precalculus

0. Fundamental Concepts of Algebra

Factoring Polynomials

Struggling with Precalculus?

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Multiple Choice

Factor the polynomial by grouping
$-x^2-5x+7x+35$

$\left(-x+7\right)\left(x+5\right)$

$\left(x+7\right)\left(x+5\right)$

$\left(x-7\right)\left(x+5\right)$

$\left(x+7\right)\left(-x+5\right)$

Verified step by step guidance

Start by grouping the terms in pairs: $-x^2 - 5x$ and $7x + 35$.

Factor out the greatest common factor from each pair. For $-x^2 - 5x$, factor out $-x$, giving $-x(x + 5)$. For $7x + 35$, factor out $7$, giving $7(x + 5)$.

Notice that both groups now contain a common binomial factor $(x + 5)$.

Factor out the common binomial $(x + 5)$ from the entire expression, resulting in $(x + 5)(-x + 7)$.

Verify the factorization by expanding $(x + 5)(-x + 7)$ to ensure it matches the original polynomial $-x^2 - 5x + 7x + 35$.

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