# Question Video: Factor Theorem with Synthetic Division Mathematics • 10th Grade

Consider the function π(π₯) = 2π₯β΄ + 10π₯Β³ + 5π₯Β² β 20π₯ + 3. Using synthetic division, find the value of π(β3). State which, if any, of (π₯ β 3) and (π₯ + 3) is a factor of π(π₯).

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### Video Transcript

consider the function π of π₯ equals two π₯ to the fourth power plus 10π₯ cubed plus five π₯ squared minus 20π₯ plus three. Using the synthetic division, find the value of π of negative three.

So here, we have our function and we were to find π of negative three. Now, we could plug negative three into π₯ and evaluate and find the answer. However, using synthetic division, we can do the exact same thing. So what we need to do is to put the negative three on the outside and then take all of the coefficients of our function: two, 10, five, negative 20, and three.

So in order to use synthetic division, we must first bring down the two. And then, we will take two times negative three, which is negative six, and put it here in the second column. And now, we will add these numbers together. 10 plus negative six is equal to four.

And now, we will repeat the process. So we need to take four times negative three, which is equal to negative 12. And now, we add five and negative 12 together, which is negative seven. Negative seven times negative three is positive 21. Negative 20 plus 21 is one. One times negative three is negative three. And lastly, three plus negative three is equal to zero.

Now, this is the most important number, the zero, because this last number is going to be our answer. Usually, itβs the remainder. But here to find the value of π of negative three, itβs just that. Therefore, zero will be our final answer.

However, we need to answer one more question. It says, state which, if any, of π₯ minus three and π₯ plus three is a factor of π of π₯.

In order to be a factor of π of π₯, we would divide it. And if it werenβt evenly, there will be no remainder. So notice weβd actually already divided by negative three from our first part. So if we did negative three for the division, that must have been π₯ equals negative three. So if we wanted to know what that would be as a factor, we will need to add three to both sides of the equation. That way, it would be on the left-hand side with the π₯. So we have π₯ plus three.

So π₯ plus three is a factor. So now, we need to know does the π₯ minus three become a factor as well for π of π₯. So if we look at π₯ minus three as a factor, we will need to set that equal to zero to figure out what π₯ to divide by. So we would need to add three to both sides of the equation so we have that π₯ equals positive three.

So we will do the same thing that we did above, but with positive three. So weβve three on the outside and then two, 10, five, negative 20, and three on the inside from π of π₯. So weβll begin by bringing the two down and our two times three is six. 10 plus six is 16. And 16 times three is 48. And five plus 48 is 53. And 53 times three is 159. And negative 20 plus 159 is 139. And 139 times three is 417. And three plus 417 is 420.

So this last piece would be the remainder. And it would need to be zero if we will divide evenly and become a factor. So the three did not work, which means the π₯ minus three factor is not one.

So the only factor that we have would be π₯ plus three. Therefore, only π₯ plus three is a factor.