Factor the polynomial by grouping. 6x3−2x2+3x−16x^3-2x^2+3x-16x3−2x2+3x−1

(2x2+1)(3x−1)\$$left(2x^2+1$$\right)\left(3x-1\right)(2x2+1)(3x−1)

Factor the polynomial by grouping. 6 x 3 − 2 x 2 + 3 x − 1 6x^3-2x^2+3x-1 6 x 3 − 2 x 2 + 3 x − 1

Welcome back, everyone. Let's take a look at this polynomial here. We're gonna factor this polynomial using the grouping method. So let's first make sure that this polynomial is in standard form, which it is. I have x cubed, x squared, x, and then just something to the zero power. So I'm good to go. So now I just separate my groups into 2 pairs. Let's take a look at these first two numbers and these last two terms over here, break them into their own groups. Now we're gonna see if we can pull out a greatest common factor out of this first term over here. So what is the 6 multiplied down to? Well, first, let's take a look at the numbers. I have 6 over here and 2 over here. So if you look at what happens here, the 6 breaks down to to 2 times 3, or you could write 3 times 2, whichever one you want. 3 times 2 times 3 powers of x over here, and then I've got 2 over here, so this just breaks down into 2 times x times x. So if you notice what happened here, what's common between them? Well, I've got a 2 that's common with a 2, I've got one power of x that's common, and I actually have another power of x that's common. So, it turns out that this whole entire 2xxx can be pulled out of that greatest common factor or this first group over here. So, I'm gonna pull out the 2 x squared to the outside of the parentheses. And what am I left with? I write 2 x squared, parentheses, 3 times x. That's what remains on the inside. And then what happens over here? I've canceled out the entire term. So what am I left with? Well, what happens is I'm not left with 0. I'm actually left with 1. So I've basically divided out this 2 x squared to the outside of the expression, and whenever you divide something out completely, you're not left with 0. There's a one that's hidden there. And if you're ever, you know, if you're if you don't believe me, think about how you would distribute this back into this expression. If you distribute the 2 x squared to the 3 x, you get 6x cubed. And if you distribute the 2 x squared into the negative one, you should get 2 negative 2x squared again. So there has to be a one that's still here. Just be really careful when you sort of divide stuff out like that. And again, if you're ever unsure, just always work backwards. So the 2x2, 3x-one. That's the first group. Let's take a look at the second one. The second one, what you'll see here is that we actually just have 3x+1-1. So if you take a look at what's happened here, we've already have we've already gotten our expressions because I can't really reduce this 3x minus 1 any further. So it turns out that they've basically can move on to my 4th step over here, which is I can now factor out the greatest common factor from both of the groups. Let's take a look at what happens. If I take 3x1 and bring it to the outside over here, I have 3x+1, That is the first term. What am I left with on the inside? Well, I have a 2x squared over here, 2x squared. And then what happens when I divide out the 3x minus 1? It's basically the exact same thing that we just saw over here. We were able to pull out a term completely to the outside of the expression. So what are you left with? It's not 0. You're left with just a factor of 1. Again, multiply it back in. 3x minus 1 into the 2x squared, and you should get back to this. And the 3x minus 1 into the just 1, you just get into you know, you just have another factor of 3x minus 1. And if you multiply this whole thing out, you should get back to your original expression over here. So this is how you factor this, and every time you pull something out completely, you always are left with a factor of just one there. Alright? So this is how you factor using the greatest common factor. Let me know if you have any questions. Thanks for watching.

Factor the polynomial by grouping. 6x3−2x2+3x−$$16x^3$-2x^2+3x-16x3$$−2x2+3x−1

(2x2+1)(3x−1)\$$left(2x^2+1$$\right)\left(3x-1\right)(2x2+1)(3x−1)

2x2(3x−$$1)2x^2$$\left(3x-1\right)2x2(3x−1)

(2x2+x)(3x−1)\$$left(2x^2+x$$\right)\left(3x-1\right)(2x2+x)(3x−1)

(2x+1)(3x2−1)\left(2x+1\right)\$$left(3x^2-1$$\right)(2x+1)(3x2−1)

0. Review of Algebra

Factoring Polynomials

College Algebra

0. Review of Algebra

Factoring Polynomials

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Multiple Choice

Factor the polynomial by grouping.
$6x^3-2x^2+3x-1$

$2x^2$\left$(3x-1$\right$)$

$$\left$(2x^2+x$\right$)$\left$(3x-1$\right$)$

$$\left$(2x+1$\right$)$\left$(3x^2-1$\right$)$

$$\left$(2x^2+1$\right$)$\left$(3x-1$\right$)$

Verified step by step guidance

Start by grouping the terms in pairs: (6x^3 - 2x^2) and (3x - 1).

Factor out the greatest common factor from each pair. For the first pair, factor out 2x^2, giving 2x^2(3x - 1). For the second pair, there is no common factor, so it remains as (3x - 1).

Notice that both groups contain the common binomial factor (3x - 1).

Factor out the common binomial factor (3x - 1) from the entire expression, resulting in (3x - 1)(2x^2 + 1).

Verify the factorization by expanding (3x - 1)(2x^2 + 1) to ensure it matches the original polynomial.

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