One zero of every polynomial is given. Use it come express the polynomial as a product of linear and also irreducible quadratic factors.$$x^4-5 x^3+7 x^2-5 x+6 ; ext zero: x=3$$

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so problem number 27 offers a polynomial X to the 4th minus five X cube. That was seven X squared, minus five X. Yet six. We're provided the zero 3 which leads the snow that we're gonna have actually a variable X minus three. You know, we deserve to start long department where we deserve to rewrite the totality pulling on young and divided. Why space given element of X minus three? So us multiply egg by ex cute since that means X cubed times X will offer us X to the fourth and then execute minus end times negative. 3 We'll get with negative three hyuk. And when us subtract fine gets us will gain these 2 to cancel out and also will have negative five add to three, i beg your pardon will give us an adverse two x cubed. And also then we lug down our continuing to be factors. So we have an unfavorable two x cubed plus seven x squared minus five x add to six. Currently we main point x through negative, too. X squared. So that's negative. 2 X squared times X will provide us an adverse two x cubed and then an adverse two X squared times Negative. Three. Well, offer us optimistic six x squared. I'm always trucked Look in ~ the 2 excuse come cancel out and then 7 minus six will offer us x squared and also we carry down our remaining determinants again We have actually X squared minus 5 x plus six and now us multiply X mine ex to obtain it to it is in x squared and then next time 2nd of three will provide us an unfavorable three eggs when we subtract Yeah, always attract We're gonna gain that X crates come cancel out. Then we have an unfavorable five x to add Rigases an unfavorable two x and we carry it on our six again No, we multiplying x by an adverse to to obtain it to be negative two x and also never the 2 times under three will be positive. Six. And when us subtract, we're going to get everything to publication out. For this reason this would certainly be the result of her long division and we writing. We have actually X minus three and execute minus two X squared add to X minus two. And now us can aspect this by grouping so us can split this top top the middle and say, Okay, well, us can aspect out one X squared the end of these two here, for this reason we have X squared on the external than we have X minus two left there and out of here. We deserve to just fact aren't one. And also we still have X minus two left. So, no, we carry out X minus 3 X minus two. Due to the fact that that's the two that are left in there and they're the very same sorts. Simply echo minus two and also then X squared to add one.

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And the ex scripless one is are irreducible, quadratic, apologetic factor. And then I uncover out.