Factor ƒ(x) into linear factors given that k is a zero. See Examp...

Question

Factor ƒ(x) into linear factors given that k is a zero. See Example 2. ƒ(x)=2x^3-3x^2-5x+6; k=1

Accepted Answer

Hey, everyone. Welcome back in this problem. We're asked to factor the following function to express it as a product of its linear factors. If four is a zero, the function we're given is F of X is equal to four, X cubed minus 35 X squared plus 71 X plus 20. We're given four answer choices A through D, all of them show polynomials of degree three that are factored using slightly different factors. Now, we're gonna start with the first piece of information we're given, we're told that four is the zero of this function. So if four is a zero of our function, then X minus four must be a factor and this is over function F of X. Now, if that doesn't quite make sense, right? A way, let's think about this. The other way. Imagine we have a polynomial and it has X minus four as a factor. If you wanted to find the zeros, you would set that function equal to zero. You could look at the factors individually. So you would set the factor equal to zero. If you said X minus four equal to zero, you get that X is equal to four. And so four is a zero. OK. So here we're just doing the opposite, we were given the zero as four. And so we're writing the factor of the function X minus four. So this tells us that our function F F can be written as X minus four multiplied by some function G of X. And now our goal is to find what that function G of X is so that we can factor it in order to find the full factored version of F M X. Now, our answer choice is A and B do not contain the factor X minus four. They have the factor X plus four. So we can actually already eliminate option A and B because they don't have this factor corresponding to the zero we were told we had now in order to find G FX, then we can divide fedex by X minus four. It's like we're isolating G FX in our equation. But now we're dividing function instead of just numbers. In order to do that. We're gonna use long division. We set up our long division. We have our line and we're gonna write our function F of X below that line or X cubed minus 35 X squared plus X was 20. And on the left, we're gonna write the factor we're dividing by which is X minus four. Now, the first step in our division is to take the first value of our function F of X under the line four X cubed and divide it by the first term in the factor on the left, which is X four X cubed divided by X gives us four X square. We're gonna write that up above and we're gonna take that for X squared and multiply it by the entire factor that we have on the left X minus four. This gives us four X Q M- squid. Now we write this below our original function but we're gonna line up the functions in columns so that the exponents that are similar are in the corresponding column. OK. So all of the X cube terms are in one column. All of the X squared terms are in one column. We're gonna subtract those functions. We have four X cube minus four X cubed which gives us zero negative 35 X squared minus negative 16 X squared give this negative 19 X squared. Now our linear term and our constant term, there's nothing being subtracted there. So we can bring those down and they stay the same. We have plus 71 X plus 20 at the end of that function. So now we have this new function negative 19 X squared plus X plus 20. And we repeat the process again, we take the first term negative 19 X squared, we divide it by the first term on the left X. This gives us negative 19 X. We take negative 19 X, we multiply it by the entire factor X minus four. This gives us negative 19 X squared plus 76 X. Can we write them down below under the function we were dealing with? And in corresponding columns, we subtract the functions negative 19 X squared minus negative 19 X squared gives a zero X M 76 X gives us negative five X. In our constant term of 20 we can bring that down again. Nothing is subtracting there. Now we have negative five X plus 20 we repeat the process again, negative five X divided by X gives us negative five, negative five, multiplied by X minus four, gives us negative five X plus 20. And when we subtract those two functions, those functions are the exact same negative five X plus 20. For both, we subtract them and we get zero, which is great. That's the remainder. And we wanted that zero. The result of our division is the function G of X and that is the function at the top of this division. OK. So above our original function that we wrote up here, piece by piece and that is four X squared minus X minus five. And so our function F of X can be written as that original factor. We had X minus four multiplied by this function G of X which is four X squared minus X minus five. Now you can always double check the result of your division and you can multiply these two together X minus four and four X squared minus 19 X minus five. Expand, simplify and you'll get back to that original function F FX. OK. If you don't get back to that original function F FX, something's gone wrong. And it's a good idea to go through and double check your work. Now, we're almost there. We've got our linear factor and we have this quadratic term. So what we wanna do is take that quadratic term and factor it as well. Now you can factor this any way you like. And there are plenty of methods to factor whatever you're most comfortable with. I'm gonna do it by breaking up our middle term. So we have this middle term of negative 19 X right now, I'm gonna break that into negative 20 X plus X. OK? So we can write this quadratic term as four X squared. Mine is 20 X love minus five. Now, we have four terms and we can factor them two at a time. OK? So our function F of X is X minus four that stays the same multiplied by. We're gonna look at the first two terms, four X squared minus 20 X. OK? We can factor four X from both of these. And when we do that, we're left with X minus five. So that's the first two terms and we move to the second two terms, we have X minus five. Oh, we can't really factor anything there. So we just have X minus five there. Now, what you'll notice is that we have this X minus five term or factor in both terms of this expression. OK? So we can actually factor that in and when you choose to break up the middle term, OK, you wanna choose values that give you a common factor like this. OK? So we have X minus four multiplied by while we have this common factor of X minus five that we can factor out. And when we do that, we're left with four X from the first term and just one in the second term. So now our function F of X is completely factored and we haven't written as F of X is equal to X minus four, multiplied by X minus five, multiplied by four X plus one. And again, if you wanted to double check, you can expand this multiply those terms together simplify and you should get that original function F FX back if we go up to our answer choices and we compare what we found, OK? We had already eliminated answer choices A and B and we see that the answer choice we found corresponds with answer D. 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. See Example 2. ƒ(x)=2x^3-3x^2-5x+6; k=1

Key Concepts

Factoring Polynomials

Zeros of a Polynomial

Synthetic Division