### Video Transcript

Liam used synthetic division to prove that four is a root of the polynomial π of π₯ equals two π₯ cubed minus nine π₯ squared plus π₯ plus 12. Using his result, factor π of π₯ into three linear factors.

So first letβs go through and understand the synthetic division that Liam used. So he took his function π of π₯ and used the coefficients and constants β two, negative nine, one, and 12 β and then he divided by four to prove it was a root. And the way that we know itβs a root is it would divide into that functions evenly.

Since there is a remainder of zero, it did go in evenly. So four would be a root. So even though we were strictly using numbers for this synthetic division, when dividing by four, weβre actually taking our functions and dividing by π₯ minus four, because think about it, if weβre dividing by π₯ equals four, if we would move four over to the left with the π₯, we would have to subtract it from both sides of the equation and we would have that π₯ minus four is equal to zero.

So π₯ minus four is one of the roots. And weβre supposed to use his result to find three linear factors. So we need to find the other two. Well, when he divided by the π₯ minus four, we have an answer. As we said zero is the remainder, negative three is the number that is constant, negative one is the coefficient with π₯, and two is the coefficient with π₯ squared.

So weβre left with two π₯ squared minus π₯ minus three equals zero. So this is whatβs left after dividing our function by four. So our last two factors will be made up from this. So we can factor this to find our last two factors. And since itβs an advanced trinomial because the leading coefficient isnβt one, we can use the slip and slide method.

So we will slip the two back with the negative three and get π₯ squared minus π₯ minus six equals zero. So we need to find two numbers that multiply to be negative six and add to be negative one; that would be negative three and positive two. Now the slide part comes in where we need to slide the two that we had slipped to the back, we slide it right underneath these.

And then we simplify. Three-halves does not simplify but two-halves does simplify. Two over two is equal to one. Now we do not leave it as π₯ minus three halves. We will take the two and bring it up with the π₯, making our other two factors two π₯ minus three and π₯ plus one.

So here are our linear factors. It doesnβt matter the order that we write them, but we have π₯ minus four, two π₯ minus three, and π₯ plus one.