Factor the polynomial.4x2−19x+124x^2-19x+124x2−19x+12

( 4 x − 3 ) ( x − 4 ) \left(4x-3\right)\left(x-4\right) ( 4 x − 3 ) ( x − 4 )

Factor the polynomial. 4 x 2 − 19 x + 12 4x^2-19x+12 4 x 2 − 19 x + 12

Hey, everyone. So let's take a look at this polynomial here. We've got 4x2minus19x plus 12, and we wanna factor it. So let's go over our factoring methods. So I can't pull out a greatest common factor here because while 4 does go into 12, it doesn't go into 19, so I don't have a common factor that works for all the terms. I also can't use grouping because I don't have 4 terms, and I also can't use a special product because although 4 x is a perfect square, 12 isn't, and it's not a perfect cube and neither is 19. It doesn't look like any of those things. So I'm gonna have to use the AC method for this. And so remember, I just want to make sure that this polynomial is in standard form. I've got something that looks like of the form ax2+bx+c And so what I'm gonna do here is I've also noticed that my a term is not equal to 1. And so how do I factor this? Well, I'm just gonna stick to the steps. The first thing I have to do is I have to build up my little table, and I'm looking for numbers, two numbers that multiply to a times c. Now in this case, what happens is it's not 12 because that's just the c term. We have to multiply a times c four times 12, and that just gets me 48. So what I have to do is list out numbers that multiply to 48, positive and negative ones. Now to save some space, one of the things you can do is if I'm looking for 48 then then both numbers either have to positive or negative. So what I can do here is having to instead of having to write 2 different rows, for 148, you can kind of just condense it by doing 148 plus and minus. And so let's do 2. So I could do 224 or negative 2 and negative 24. So this is just a kind of like a way to make it a little bit more compact. Compact. Let's keep going. What about 3? Does 3 go into 48? It does because I could do 316. So that means I'm gonna just do plus or minus 3, plus or minus 16. What about 4? 4 times 12 is equal to 48, so I could do plus -four, plus -twelve. What about 6? Well, 6 goes into 8 as well because 6 times 8 oh, sorry, it's 48. 6 times 8 is 48, so plus or minus. Okay. And then if I just go if I keep going, I'm gonna get to 8, but I've already covered that before, so I'm basically done here. So these are all just my unique pairs that multiply to 48. That brings us to the 3rd step, which is now we have to figure out which one of these numbers add to the b term. The b term in this case is equal to negative 19. Not positive 19, it's negative. Alright? So here's what happens. If I do 148, I get 49. I get 49 here. And if I do negative one and negative forty eight, I get negative 49. So in other words, this will add up to either positive or negative 49. So I'm just gonna do plus or minus, But neither one of those are our answers, so it doesn't matter. Right? What about 224? 224 will get me, 48 sorry. I'm sorry. 20 26 and either plus or minus. So that's not what I'm looking for either, so that doesn't work. What about 316? 316 will get me 19 or negative 19, and that's actually exactly what I'm looking for here. So, it turns out that I can basically just stop here because these combinations aren't going to work because I found my two numbers. What What are the two numbers that get me to -nineteen? Well, it's gonna be the negatives of these two numbers, -three and -sixteen. So -three and -sixteen, that's basically what my p and q terms are. So that's all there's to it. You're basically done here. Now now remember, instead of having to just plug instead of just plugging these numbers into my x+p andx+q, we can't do that because a is not equal to 1. Because, so what we have to do instead was we have to actually split out this term into a 4 term expression, and then we're going to use the grouping method. So what this turns out to is this turns out to 4x2 minuteus and I do the minus the 3x. So in other words, that's my px, and then the qx over here is -16x plus 12. So basically, just splitting out the 19 into 316 both, you know, negatives and stuff. Right? So now what I can do is because this is a 4 term expression in which I do see some common factors, now I can use the grouping method for this. Alright? So let's take a look here. What I've got is I'm gonna just sort of group together the first two terms and then the last two terms over here, see if I can pull out a greatest common factor out of each one of them. So what happens? I have a 4x2 and a 3x, and I'm gonna skip some steps here, but, basically, 3 doesn't go into 4. But I do notice that there's at least one power of x that's common between both of them. So I pull out that x to the outside, and then what's left is on the inside is I have 4x minus 3. Right? So you basically cross out a factor of x from both of these things. Okay. So then what happens over here? Well, for the second group, I've got negative 16x plus 12. So both of those, while those are not multiples of each other, I do know that 4 goes into both of these things over here. So when I factor these out, this actually turns out to be, negative 4 times 4 x plus and then I've got 4 times 3. So in other words sorry. I'm gonna do 4 times x. So in other words, what I've seen here is that in this group, I actually can pull out a 4 that's common to the outside. So I'm gonna do that and cancel this out, and what I end up with here is this expression is x4x-three, and then I do plus 4, and then I have negative 4x plus 3. Alright? Now remember, usually in these grouping problems, what what you end up with is you end up with the same exact expression over here, but here I didn't. Here what happens is I have 4x minus 3, and here I have negative 4x plus 3. So what gives? How do I actually solve this? Well, it turns out that these things are basically just the same thing. The only thing that's different is the is, like, a minus sign in both of the terms. So what I can do here is I can basically just pull out a negative one factor because this is really negative one times four times x, And this negative three over here, it's kind of weird, but I can basically pull out a negative one times negative one times three. So what happens here is that whenever you have 2 expressions and you don't see the same sign, it's usually because you can pull out one one factor of -one to the outside. And here's what happens. This really just becomes x times 4x-three minus 4, and this is 4x minus 3. If you actually plug in these two expressions, if you do 4 into negative 4x plus 3 and negative 44x minus 3, they actually mean the exact same thing. You can even distribute them out and see for yourself that these 2 are equal to each other. So these 2 are equal. Alright. So here's the so that's what happens there. If you're ever off by a sign, it just means that you can pull out an a negative one. And so now what happens is now I do notice that I do have the same expression in the entire throughout the entire both groups, so I can pull that out to the outside as well. And so this just becomes 4x minus 3 in parentheses, and then I have x minus 4. And if you go ahead and foil this out, you should get back to your original expression over here. Alright. So hopefully, that made sense. This step is pretty tricky here, but you'll see that when you do this and you end up with stuff that, you know, where you're like a sign off, it's usually because there's, like, a negative one that you can extract to the outside. Alright? So that's it for this one, folks. Let me know if you have any questions.

Factor the polynomial. 4 x 2 − 19 x + 12 4x^2-19x+12 4 x 2 − 19 x + 12

( 4 x − 3 ) ( x − 4 ) \left(4x-3\right)\left(x-4\right) ( 4 x − 3 ) ( x − 4 )

( 4 x − 6 ) ( x − 2 ) \left(4x-6\right)\left(x-2\right) ( 4 x − 6 ) ( x − 2 )

( 2 x − 3 ) ( 2 x − 4 ) \left(2x-3\right)\left(2x-4\right) ( 2 x − 3 ) ( 2 x − 4 )

( 2 x − 6 ) ( 2 x − 2 ) \left(2x-6\right)\left(2x-2\right) ( 2 x − 6 ) ( 2 x − 2 )

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.
$4x^2-19x+12$

$\left(4x-3\right)\left(x-4\right)$

$\left(4x-6\right)\left(x-2\right)$

$\left(2x-3\right)\left(2x-4\right)$

$\left(2x-6\right)\left(2x-2\right)$

Verified step by step guidance

Identify the polynomial to be factored: $4x^2 - 19x + 12$. This is a quadratic polynomial in the form $ax^2 + bx + c$.

Look for two numbers that multiply to $a \times c = 4 \times 12 = 48$ and add up to $b = -19$. These numbers will help split the middle term.

The numbers that satisfy these conditions are $-3$ and $-16$ because $-3 \times -16 = 48$ and $-3 + (-16) = -19$.

Rewrite the middle term $-19x$ using the numbers found: $4x^2 - 3x - 16x + 12$.

Factor by grouping: Group the terms into pairs, $(4x^2 - 3x)$ and $(-16x + 12)$, and factor each pair separately to find the common factors, leading to the final factored form.

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