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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.
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Using his result, factor π of π₯ into three linear factors.
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So first letβs go through and understand the synthetic division that Liam used.
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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.
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And the way that we know itβs a root is it would divide into that functions evenly.
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Since there is a remainder of zero, it did go in evenly.
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So four would be a root.
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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.
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So π₯ minus four is one of the roots.
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And weβre supposed to use his result to find three linear factors.
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So we need to find the other two.
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Well, when he divided by the π₯ minus four, we have an answer.
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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.
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So weβre left with two π₯ squared minus π₯ minus three equals zero.
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So this is whatβs left after dividing our function by four.
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So our last two factors will be made up from this.
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So we can factor this to find our last two factors.
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And since itβs an advanced trinomial because the leading coefficient isnβt one, we can use the slip and slide method.
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So we will slip the two back with the negative three and get π₯ squared minus π₯ minus six equals zero.
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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.
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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.
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And then we simplify.
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Three-halves does not simplify but two-halves does simplify.
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Two over two is equal to one.
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Now we do not leave it as π₯ minus three halves.
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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.
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So here are our linear factors.
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It doesnβt matter the order that we write them, but we have π₯ minus four, two π₯ minus three, and π₯ plus one.