Question Video: Writing Polynomials as a Product of Linear and Irreducible Quadratic Factors and Listing All Zeros Mathematics

Consider 𝑔(π‘₯) = π‘₯⁴ βˆ’ 7π‘₯Β³ + 11π‘₯Β² βˆ’ 41π‘₯ + 180. Write 𝑔(π‘₯) as the product of linear and irreducible quadratic factors. Write 𝑔(π‘₯) as the product of linear factors. List all zeros of 𝑔(π‘₯).

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Video Transcript

Consider 𝑔 of π‘₯ equals π‘₯ to the fourth power minus seven π‘₯ cubed plus 11π‘₯ squared minus 41π‘₯ plus 180. Write 𝑔 of π‘₯ as a product of linear and irreducible quadratic factors. Write 𝑔 of π‘₯ as a product of linear factors. And list all zeroes of 𝑔 of π‘₯.

So, the first part of this question has asked us to write the expression π‘₯ to the fourth power minus seven π‘₯ cubed plus 11π‘₯ squared minus 41π‘₯ plus 180 as the product of linear and irreducible quadratic factors. The highest power of π‘₯ in our function is π‘₯ to the fourth power. So, we know that if we have linear and irreducible quadratic factors, 𝑔 of π‘₯ will be written as π‘₯ plus π‘Ž times π‘₯ plus 𝑏 times π‘₯ squared plus 𝑐π‘₯ plus 𝑑, where π‘Ž, 𝑏, 𝑐, and 𝑑 are real constants. So, there are two ways that we can answer this problem. The first would be to distribute the set of parentheses and equate coefficients.

It can be quite a long way to do it, so instead we’ll recall the factor theorem. And this says that if π‘₯ minus π‘Ž is a factor of some function 𝑓 of π‘₯, then 𝑓 of π‘Ž will be equal to zero. Well, we can find π‘Ž using a little bit of trial and error. We know π‘Ž itself will be a factor of 180. The first few factors of 180 are one, two, three, four, five, and so on. So, what we’re going to do is try substituting some of these factors into our function. And if we get zero, we know that we found a factor. Let’s begin by trying 𝑔 of two. That’s two to the fourth power minus seven times two cubed plus 11 times two squared minus 41 times two plus 180. Well, that’s equal to 102, so clearly π‘₯ minus two is not a factor.

Next, we’ll try π‘₯ equals four. This time we get four to the fourth power minus seven times four cubed plus 11 times four squared minus 41 times four plus 180. Well, this time that is equal to zero. And that means π‘₯ minus four is itself a factor of 𝑔 of π‘₯. Similarly, if we try π‘₯ equals five, we also get zero. So, π‘₯ minus five must also be a factor of 𝑔 of π‘₯. So, with this stage, we know 𝑔 of π‘₯ can be written as π‘₯ minus four times π‘₯ minus five times π‘₯ squared plus 𝑐π‘₯ plus 𝑑. And there are two ways again that we could work out this quadratic; we could distribute our parentheses and equate coefficients.

Now, alternatively, if we multiply π‘₯ minus four by π‘₯ minus five, we get π‘₯ squared minus nine π‘₯ plus 20. And we can now use polynomial long division to work out the other factor, the other quadratic. π‘₯ to the fourth power divided by π‘₯ squared is π‘₯ squared. Then, we multiply π‘₯ squared minus nine π‘₯ plus 20 by π‘₯ squared. And we get π‘₯ to the fourth power minus nine π‘₯ cubed plus 20π‘₯ squared. We then subtract each of these terms. Negative seven π‘₯ cubed minus negative nine π‘₯ cubed is two π‘₯ cubed. 11π‘₯ squared minus 20π‘₯ squared is negative nine π‘₯ squared. And we bring down the next term. Then, two π‘₯ cubed divided by π‘₯ squared is two π‘₯. We multiply two π‘₯ by π‘₯ squared, negative nine π‘₯, and 20 to get two π‘₯ cubed minus 18π‘₯ squared plus 40π‘₯. We subtract again. And that gives us nine π‘₯ squared minus 81π‘₯.

Once again, we bring down that extra term, and this time we divide nine π‘₯ squared by π‘₯ squared and we get nine. And multiplying nine by π‘₯ squared, negative nine π‘₯, and 20 gives us nine π‘₯ squared minus 81π‘₯ plus 180. And this time, when we subtract, we get zero. We get no remainder, which is ideal because we knew that π‘₯ minus four and π‘₯ minus five are factors of 𝑔 of π‘₯. We’re told that this quadratic is irreducible, so we finished. 𝑔 of π‘₯ is equal to π‘₯ minus four times π‘₯ minus five times π‘₯ squared plus two π‘₯ plus nine.

Now, the second part of this question asked us to write 𝑔 of π‘₯ as the product of linear factors. Normally, we’d be looking to factorize π‘₯ squared plus two π‘₯ plus nine. But we’re told that’s not possible. So instead, we’re going to use completing the square to find the roots of π‘₯ squared plus two π‘₯ plus nine equals zero. Once we’ve done that, we’ll be able to put this back into the expression. We begin by halving the coefficient of π‘₯. That’s one, so we get π‘₯ plus one all squared. We then take away one, which is one squared, and bring the rest of our equation down. So, we have π‘₯ plus one all squared plus eight equals zero. We then subtract eight from both sides. And we get π‘₯ plus one squared equals negative eight and find the positive and negative square root of negative eight. Then, we subtract one from both sides, so we see the π‘₯ is equal to negative one plus or minus the square root of negative eight.

We’re going to rewrite the square root of negative eight as the square root of eight times negative one. And then, that’s equal to the square root of eight times the square root of negative one. Well, we can simplify the square root of eight to two root two. And of course, the square root of negative one is 𝑖. So, we find π‘₯ is equal to negative one plus or minus two times the square root of two 𝑖. We can now put this back into the original expression. And this time we find that 𝑔 of π‘₯ is equal to π‘₯ minus four times π‘₯ minus five times π‘₯ minus negative one plus two root two 𝑖 times π‘₯ minus negative one minus two root two 𝑖. By distributing the parentheses, we find that these final two terms become π‘₯ plus one minus two root two 𝑖 and π‘₯ plus one plus two root two 𝑖.

The final part of this question asked us to list all zeroes of 𝑔 of π‘₯. In other words, solutions to the equation 𝑔 of π‘₯ equals zero. Now, for this expression to be equal to zero, either π‘₯ minus four is equal to zero, π‘₯ minus five is equal to zero, π‘₯ plus one minus two root two 𝑖 is equal to zero, or π‘₯ plus one plus two root two 𝑖 is equal to zero. Now, of course, for π‘₯ minus four to be equal to zero, π‘₯ would be equal to four. If π‘₯ minus five is equal to zero, π‘₯ is equal to five. If π‘₯ plus one minus two root two 𝑖 was equal to zero, π‘₯ would be equal to negative one plus two root two 𝑖. And finally, if π‘₯ plus one plus two root two 𝑖 was equal to zero, π‘₯ would be equal to negative one minus two root two 𝑖.

And so, we have the zeroes of 𝑔 of π‘₯. They are four, five, negative one plus two root two 𝑖, and negative one minus two root two 𝑖.

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