Use synthetic division to show that 5 is a solution of x^4−4x^3−9...

Question

Use synthetic division to show that 5 is a solution of x^4−4x^3−9x^2+16x+20=0. Then solve the polynomial equation.

Accepted Answer

Welcome back. I am so glad you're here. We are given this polynomial equation X. To the fourth minus five X. Cubed minus 49 X squared plus 125 X plus 600 equal zero. And we are asked to see if eight is a solution using synthetic division. Alright. We recall from previous lessons on synthetic division that we're going to take that eight, write it off to the side here and then we're going to copy down all of the coefficients and the constant term from our polynomial equation that's being divided. So starting with this X to the fourth, we always want to start with the highest degree and work our way down, not missing a single degree of X. So for X to the fourth, the coefficient term here is one. So we write down one and then just going down in descending order we have negative five negative 125 and 600. And now we are going to follow the steps for synthetic division. The first step only happens one time. We are going to bring down that first term, so bring down the one and write it here. Now the next step is step one of what repeats. So all right, Step one again. And this step one that repeats. Is to multiply we're going to take the eight times the one and write the answer here eight times one is eight. And step two which will be repeated is to add we take negative five plus eight. And that gives us three. Okay, back to step one. Multiplying eight times three is 24. Now adding again negative 49 plus 24 is negative 25. Back to multiplying eight times negative 25 is negative 200. Back to adding 125 -200 or plus a -200 is going to be negative 75. At multiplying again eight times negative 75 is negative 600 adding 600 plus a negative 600 is zero. Alright, once we've completed this synthetic division, we look at this last term here and we ask ourselves is this equal to zero? And yes it is. That means we have no remainder and it means that eight is a solution for this polynomial equation. Yay we've got one of them now. I gotta get the rest. Alright we recall that when we do synthetic division we come up with a new polynomial equation here. This first term. This one is the leading coefficient for a new polynomial equation, that is one degree less than our previous polynomial equation. The degree of the previous polynomial equation was four. So now this one is the coefficient for a polynomial with a degree of X to the third, one less than X to the fourth. So we'll just keep copying down these terms. They become our new coefficients going down and descending order. So one less than X to the third is going to be X squared and bring down this three. So it's plus three, X squared. Bring down this minus 25 down one degree from X squared will be X. And then one degree down from regular X is X to the zero which is one. So this negative 75 is our new constant term and this is all equal to zero because the original polynomial equation was equal to zero. And now we are going to solve for X. We can use factoring by grouping and group these first two terms together and the second two terms together. Ask ourselves, is there anything we can factor out of the first two terms? Yes, they both have an X squared in common and dividing both terms by X squared will leave us with X plus three. Looking at the second set of two terms. Anything that they have in common? Yes, a negative 25. We can factor that out and once we divide both terms by negative 25 we're left with X plus three. And this is great. These two sets have a common factor here. They both have a factor of X plus three. So we can factor X plus three right out and X plus three is being multiplied by both X squared and negative 25. And if we look at that X squared minus 25 we notice that we have a difference of squares, X squared and 25 is also a square. So we recall from previous lessons that we can factor a difference of squares by taking the square root of both of those squares, X squared. The square root of that is X. So we put an X in the first position in both sets of parentheses and the 25. The square root of 25 is five. That goes in the second position for both sets of parentheses. And by definition one will be plus and one will be minus. Ye. We have factored this, that's as factored as it's going to get. So we can set each of these three factors equal to zero. We have X plus three equals zero. X plus five equals zero and x minus five equals zero. Well we'll get those exes by themselves for all three of these equations. And when we take three from both sides for X plus three equals zero we get a solution of X equals negative three. There's one when we subtract five from both sides for X plus five equals zero we get x equals negative five. There's another one. And for x minus five when we add five to both sides we get a solution of X equals five. Well there's our other three solutions. In addition to that eight we have already found looking at our answer choices. This matches with answer choice D While done, we'll catch you on the next one

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Use synthetic division to show that 5 is a solution of x^4−4x^3−9x^2+16x+20=0. Then solve the polynomial equation.

Dividing Polynomials practice set