Factor $f with hook x$

Question

Factor

ƒ (x)

into linear factors given that k is a zero.

ƒ (x) = x^{4} + 2 x^{3} - 7 x^{2} - 20 x - 12; k = - 2

(multiplicity

2

)

Accepted Answer

Hey, everyone in this problem, we're asked to factor the following function to express it as a product of its linear factors. If negative three with the multiplicity of two is a zero, the function we're given is F of X is equal to X to the exponent four plus five, X cubed minus 17, X squared minus 129 X minus 180. We're given four answer choices A through D, they're all polynomial functions. OK? And they each have different factors. OK? To represent that polynomial, we're gonna start with what we were given. So we were told that negative three is a zero of this function. So if negative three is a zero, then X minus negative three, which we can write us at plus three is a factor of our function. So we have the zero, negative three that tells us that X plus three is a factor. OK? And we can imagine we're working kind of backwards here. But you can imagine if you had a polynomial, you factored it and X plus three was one of the factors. And you tried to solve for the zeros, you would set the equation equal to zero. And then you would look at each factor. Now, if you have the factor X plus three, you said that equal to zero, you get that X is equal to negative three. OK? So you get that zero of negative three. So here we're just doing the opposite, we have the zero. So now we're writing out the factor. OK. So X plus three is a factor and we know it has a multiplicity of two. Now recall that the multiplicity tells us how many times that factor appears in our polynomial. OK. So we have a multiplicity of two. So this factor appears twice. And so we can write our function F of X as X plus three squared. OK. That's squared comes from a multiplicity of two. We have two of these factors of X plus three multiplied by GEO back some function. Now, the goal here is to find G of X so that we can figure out the rest of the factors of our polynomial F and we find G of X, we factor it and we have the factor version of F of X. Now, in order to get G of X, what we want to do is divide F of X by X plus three squared. OK. We have the G of X is going to be equal to F of X divided by X plus three squared. Now, in order to do this division, OK, we're dividing function So we're gonna use long division, we want to write the denominator in its expanded form in order to do that. OK? So we're gonna write X plus three squared in its expanded form. OK? So X plus three squared again, X plus three multiplied by X plus three. You can expand this any way you'd like. You can use the foil method or another method that works for you. But the key is that we have to multiply every term in the first factor by every term in the second factor. And when we do that, we get X squared plus six X plus nine. OK? So now we're gonna set up our long division where we're dividing F of X by X plus three all squared. We're gonna draw our long division kind of line here and underneath it, we're gonna put the function F of X which is X to the exponent four plus five X Q M 17 X squared M 129 X minus 180. And on the left hand side of our division, we're gonna write what we're dividing by which is X squared plus six X plus nine. So now we have our division set up and we need to actually perform the division. We're gonna start with the term on the left hand side underneath our division line. OK? So the first term of our polynomial F FX, which is X to the exponent four, we're gonna divide that by the first term of the factor on the left hand side, X squared X to the exponent four divided by X squared gives us X squared. We're gonna write that value above. Then we're gonna take that X squared, we just found and multiply it by the entire polynomial on the left-hand side, X squared plus six X plus nine. OK. X squared multiplied by X squared gives us X to the exponent for X squared multiplied by six X gives us six X Q X squared multiplied by nine, gives us nine X squared. And we write that below our original function F of X. And we're gonna line up corresponding terms in columns. OK. So all of the X to the exponent four terms are written in one column, all of the X cube terms in one column. And so now we have these columns, OK? Right now we have two rows, our original function in this new function. And we're gonna track the two rows. We have X to the exponent four minus X to the exponent four that gives us zero. We have five X cubed minus six X cubed. So we get negative X cubed, negative 17 X squared minus nine X squared gives us negative 26 X squared. And then the last two terms, we're not subtracting anything and those are gonna get moved down. So we have minus 129 X minus 180. Now we have this new function down below and we repeat the same process. We take the first term negative X cubed, we divide it by the first term on the left hand side X squared that gives us negative X. We write it above. We take that negative X, we multiply it by the entire expression on the left-hand side, negative X multiplied by X squared plus six X plus nine. This gives us negative execute M six X squared minus minus. We write that down below underneath the function we were working with and lining up those columns again. And now we subtract X cubed minus X cued gives us zero negative 26 X squared minus negative six X squared gives us negative X word, negative 129 X M negative nine X gives us negative 120 X. In our final term. Again, we have nothing to subtract. So we just bring that down and we have minus 180 at the end of our function. Now again, we have this new function. We repeat the process negative 20 X squared divided by X squared gives us negative 20. We write it up above. We take that value negative 20. We multiply it by X squared plus six X plus nine. OK. Doing that gives us negative 20 X squared minus 120 S minus 180. We write that below the function we were dealing with and we're gonna subtract. Now, you'll notice that this is the exact same function. We have negative 20 X squared minus 120 X minus 80. OK? And we're subtracting the entire function. That's exactly the same. And so we're gonna get zero. OK. So we have a remainder of zero, which is great. That means that we're gonna factor this polynomial nicely. That's exactly what we wanted. And the function above our division above that original function is that function G M X, that's the result of our division. OK. So we found that that is X squared minus X minus 20. And so our function F F can be written as X plus three squared multiplied by this function that we've just found X squared minus X minus 20. Now, we have to be a little bit careful. OK. The question is asking us to factor this polynomial completely. So we still need to factor this second part a little bit more. Yeah, we have a quadratic term there. So let's go ahead and factor now X plus three squared. That's factored. We're gonna leave that one, X squared po uh minus X minus can be written as X squared plus four, X minus five, X minus 20. OK. So these two expressions are equivalent, we're just breaking up that middle term into two parts. OK? So that we can factor, we have our X plus three squared. Now we're gonna factor this first two terms and then the second two terms. OK. So in the first two terms, we have X squared plus four X, we can factor O X right. And we're left with X plus four. In the second two terms, we can factor in negative five. You are left with X plus four. All right. Now, we have this common factor of X plus four in each of these terms in the second part of our equation. So we can take out that X plus four. And when we do that, we're left with X minus five. So we factored X squared minus X minus 20 into X plus four multiplied by X minus five. OK. And you can factor any way that you'd like. OK. I showed one way here. But if you have a different method that works better for you, then go ahead and use that. And now we have our final factored function F of X. We have that F of X is equal to X plus three squared multiplied by X plus four, multiplied by X minus five. We compare this to our answer choices. We see that this corresponds with answer choice. B thanks everyone for watching. I hope this video helped see you in the next one.

Factor $ƒ (x)$ into linear factors given that k is a zero. $ƒ (x) = x^{4} + 2 x^{3} - 7 x^{2} - 20 x - 12; k = - 2$ (multiplicity $2$ )

Key Concepts

Polynomial Zeros and Multiplicity

Polynomial Division (Synthetic or Long Division)

Factoring Polynomials into Linear Factors

Factor ƒ(x)ƒ(x) into linear factors given that k is a zero. ƒ(x)=x4+2x3−7x2−20x−12; k=−2ƒ(x)=x^4+2x^3-7x^2-20x-12;\text{ }k=-2 (multiplicity 22)

Key Concepts

Polynomial Zeros and Multiplicity

Polynomial Division (Synthetic or Long Division)

Factoring Polynomials into Linear Factors

Factor $ƒ (x)$ into linear factors given that k is a zero. $ƒ (x) = x^{4} + 2 x^{3} - 7 x^{2} - 20 x - 12; k = - 2$ (multiplicity $2$ )