Video: Finding the Set of Zeros of a Polynomial Function

Find, by factoring, the zeros of the function 𝑓(π‘₯) = 6π‘₯⁴ + 33π‘₯Β³ βˆ’ 63π‘₯Β².

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Video Transcript

Find, by factoring, the zeros of the function 𝑓 of π‘₯ equals six π‘₯ to the fourth power plus 33π‘₯ cubed minus 63π‘₯ squared.

Let’s begin by recalling what is meant by the zeros of a function. These are the values of the variable, in this case π‘₯, such that the function itself is equal to zero. So we’re looking for the values of π‘₯ that make 𝑓 of π‘₯ equal to zero. We therefore need to solve the quartic equation six π‘₯ to the fourth power plus 33π‘₯ cubed minus 63π‘₯ squared equals zero. We’re told in the question that this can be done by factoring. So let’s begin by saying if we can identify any common factors in our three terms.

Looking at the coefficients, first of all, or at least their absolute values, so that’s six, 33, and 63, we see that these share a common factor of three. So we can factor this out. We then see that we have π‘₯ to the fourth power, π‘₯ cubed, and π‘₯ squared. So the lowest power of π‘₯, which appears in all three terms, is π‘₯ squared. We can therefore take three π‘₯ squared out as a common factor.

To get six π‘₯ to the fourth power, we need to multiply three π‘₯ squared by two π‘₯ squared. So that’s the first term inside our parentheses. To get 33π‘₯ cubed, we need to multiply three π‘₯ squared by 11π‘₯. And then, to get negative 63π‘₯ squared, we need to multiply three π‘₯ squared by negative 21. So we now have our equation in the partially factored form three π‘₯ squared multiplied by two π‘₯ squared plus 11π‘₯ minus 21.

Next, we need to consider how we can factor this quadratic, two π‘₯ squared plus 11π‘₯ minus 21. And as it has a leading coefficient, which is not equal to one, it has a leading coefficient of two, we need to use factoring by grouping. We look for two numbers which sum to the coefficient of π‘₯, that’s positive 11, and which multiply to the product of the coefficient of π‘₯ squared and the constant term. That’s two times negative 21, which is negative 42. We can list the factor pairs of 42. They are one and 42, two and 21, three and 14, and six and seven. In order to make negative 42 then, we require that our factors have different signs. And we see that if we need them to sum to 11, then if we make three negative but keep 14 positive, then the factor pair negative three and 14 does indeed sum to 11 and multiplies to negative 42.

Now, the next step may look a little strange if you’ve forgotten some of the detail of this method. But what we do is we take that positive 11π‘₯. And we rewrite it as the sum of positive 14π‘₯ and negative three π‘₯. We then divide this new four-term quadratic expression in half. And we factor each pair of terms separately. From the first pair, two π‘₯ squared and 14π‘₯, we can take a common factor of two π‘₯, giving us two π‘₯ multiplied by π‘₯ plus seven. And from the second pair, negative three π‘₯ minus 21, we can take a common factor of negative three, giving us negative three multiplied by π‘₯ plus seven.

We then notice that the two parts of this quadratic share a common factor of π‘₯ plus seven. So we can factor by π‘₯ plus seven. We’re then left with two π‘₯ for the first term and negative three for the second. So our second factor is two π‘₯ minus three. We can of course confirm that if we were to redistribute these parentheses and simplify, we’d get the quadratic two π‘₯ squared plus 11π‘₯ minus 21, which we started off with. Our quartic equation is now in a fully factored form. So we can solve.

To do so, we take each of our factors and set them each equal to zero. We then solve each of the resulting equations. We have three π‘₯ squared equals zero. And dividing by three and then taking the square root of each side of this equation gives π‘₯ equals zero. Our second solution is π‘₯ plus seven equals zero. And subtracting seven from each side of this equation gives π‘₯ equals negative seven. And our third equation is two π‘₯ minus three equals zero. Adding three to both sides and then dividing by two gives the solution π‘₯ is equal to three over two. So we found that there are three solutions to this quartic equation and hence three values of π‘₯, which are considered the zeros of this function. π‘₯ equals zero, π‘₯ equals negative seven, and π‘₯ equals three over two.

We can of course check our work by substituting each root individually back into the equation 𝑓 of π‘₯. For example, 𝑓 of zero is equal to six multiplied by zero to the fourth power plus 33 multiplied by zero cubed minus 63 multiplied by zero squared. That’s zero plus zero minus zero, which is indeed equal to zero, telling us that the value zero is a zero of the function 𝑓 of π‘₯.

So by fully factoring this quartic equation and then setting each factor individually equal to zero, we’ve found that the zeros of the function 𝑓 of π‘₯ are negative seven, zero, and three over two.

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