Video: Finding Zeros of Polynomials When One Is Given Using Synthetic Substitution and Depressed Polynomials

Consider the function π‘˜(π‘₯) = βˆ’5π‘₯⁴ + 2π‘₯Β³ βˆ’ 30π‘₯Β² βˆ’ 88π‘₯ + 40. Given that one zero of π‘˜(π‘₯) is 1 βˆ’ 3𝑖, find all zeros of π‘˜(π‘₯) using synthetic division. Write the linear factorization of π‘˜(π‘₯).

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Video Transcript

Consider the function π‘˜ of π‘₯ equals negative five π‘₯ to the fourth power plus two π‘₯ cubed minus 30π‘₯ squared minus 88π‘₯ plus 40. Given that one zero of π‘˜ of π‘₯ is one minus three 𝑖, find all the zeros of π‘˜ of π‘₯ using synthetic division.

To use synthetic division, we first need to make sure that the exponents are given in descending order, which they are here, π‘₯ to the fourth, π‘₯ cubed, π‘₯ squared, and then π‘₯ to the first power followed by the constant. From there, we’ll copy down the coefficients, keeping the order the same as they were given in the function. Since we’re already given one zero of this function, we’ll use one minus three 𝑖 to divide the rest of the function by.

In synthetic division, we bring down the first coefficient negative five. From there, we need to multiply one minus three 𝑖 by negative five. Negative five times one equals negative five. Negative five times negative three 𝑖 equals positive 15𝑖. And that value we put directly underneath the two. And then, we add the values from the second column together, two plus negative five plus 15𝑖. Two plus negative five equals negative three, and we bring down the 15𝑖.

And then, the process begins again by multiplying one minus three 𝑖 times negative three plus 15𝑖. Negative three times one equals negative three. Negative three times negative three 𝑖 equals positive nine 𝑖. 15𝑖 times one equals positive 15𝑖. 15𝑖 times negative three 𝑖 equals negative 45𝑖 squared. We know that 𝑖 squared equals negative one. If we substitute negative one in for 𝑖 squared, we have negative 45 times negative one, which equals positive 45.

We can combine like terms. Negative three plus 45 equals 42. We can combine nine 𝑖 and 15𝑖, which gives us 24𝑖. In our third column, we’ll then add 42 plus 24𝑖. Adding column three together, negative 30 plus 42 plus 24𝑖 gives us 12 plus 24𝑖. Repeating the process again, we need to multiply one minus three 𝑖 times 12 plus 24𝑖. We’ll need to expand these two expressions. One times 12 equals 12. One times 24𝑖 equals 24𝑖. Negative three 𝑖 times 12 equals negative 36𝑖 equals negative 72𝑖 squared.

Again, we’ll need to substitute negative one in for 𝑖 squared. Negative 72 times negative one equals positive 72. 72 plus 12 equals 84, and 24𝑖 minus 36𝑖, which gives us negative 12𝑖. We now need to add negative 88 plus 84 minus 12𝑖, which gives us negative four minus 12𝑖. Next, we multiply one minus three 𝑖 by negative four minus 12𝑖. One times negative four equals negative four. One times negative 12𝑖 equals negative 12𝑖. Negative three 𝑖 times negative four equals positive 12𝑖. Negative three 𝑖 times negative 12𝑖 equals positive 36𝑖 squared.

We know that 𝑖 squared equals negative one. Positive 36 times negative one equals negative 36. We can combine like terms. Negative four plus negative 36 equals negative 40. And negative 12𝑖 plus 12𝑖 cancels out. So, we put negative 40 into our fifth column. 40 plus negative 40 equals zero. This zero in the last column confirms that one minus three 𝑖 is, in fact, a zero of this function. But what do we do now?

To continue with synthetic division, we’ll need an additional zero of this function. Whenever we have an imaginary zero like this one, one minus three 𝑖, they come in pairs. If one minus three 𝑖 is a zero of this function, one plus three 𝑖 will also be a zero of the function. We can use one plus three 𝑖 to divide with synthetic division again. We’ll follow the same procedure and in the first column bring down the negative five.

We multiply one plus three 𝑖 by negative five, which is negative five minus 15𝑖. Add column two together. The positive 15𝑖 and negative 15𝑖 cancel out. And we’re left with negative eight. Now we’re multiplying one plus three 𝑖 by negative eight. Negative eight times one plus three 𝑖 equals negative eight minus 24𝑖. Adding the terms in the third column, we get four. 12 minus eight equals four. 24𝑖 minus 24𝑖 equals zero, and again one plus three 𝑖 times four, which gives us four plus 12𝑖. Adding column four together gives us zero and confirms that one plus three 𝑖 is a zero of our function.

Now we’ll use these three coefficients to help us find the remaining zeros. After finding two of our zeros, the corresponding degree of the exponent should be two here. So, we now have remaining negative five π‘₯ squared minus eight π‘₯ plus four. To find the remaining zeros, we can factor this equation. Negative five π‘₯ times π‘₯ equals negative five π‘₯ squared, we know that we’ll need values that multiply together to equal positive four.

We can use one and four, negative one and negative four, two and two, or negative two and negative two. We also know that our middle term must be negative eight. If we use positive two, we’ll be multiplying negative five π‘₯ plus two, which gives us negative 10π‘₯. And then, we’ll multiply two times π‘₯, which gives us positive two π‘₯. Negative 10π‘₯ plus two π‘₯ equals negative eight π‘₯. The factored form is negative five π‘₯ plus two times π‘₯ plus two.

To find our zeros, we’ll need to set each factor equal to zero, zero equals negative five π‘₯ plus two and zero equals π‘₯ plus two. Starting on the left, zero equals negative five π‘₯ plus two. We subtract two from both sides of the equation. Negative two equals negative five π‘₯. Then, divide both sides of the equation by negative five. Negative two divided by negative five equals positive two-fifths. π‘₯ equals positive two-fifths.

And now on the right side, solving for zero equals π‘₯ plus two. Subtract two from both sides of the equation. And we have negative two equals π‘₯, more commonly written as π‘₯ equals negative two. We have a zero at π‘₯ equals negative two, at π‘₯ equals two-fifths, at π‘₯ equals one plus three 𝑖, and at π‘₯ equals one minus three 𝑖. And now we need to write the linear factorization of π‘˜ of π‘₯.

We can start where we factored the first two real roots. We can start with π‘˜ of π‘₯ equals negative five π‘₯ plus two times π‘₯ plus two. We don’t really want this leading coefficient to be negative, so we pull out a negative one and change our signs inside the parentheses. We’ll have negative five π‘₯ minus two times π‘₯ plus two. We also know that π‘₯ equals one minus three is a zero of our function. If we subtract negative π‘₯ from both sides of this equation, we’ll have the factor zero equals negative π‘₯ plus one minus three 𝑖.

But remember, we don’t want this π‘₯-value to be negative. So, we can multiply through by negative one, and we have zero equals π‘₯ minus one plus three 𝑖, a third factor π‘₯ minus one plus three 𝑖. And we need to repeat this step for the zero π‘₯ equals one plus three 𝑖. Subtract π‘₯ from both sides. Zero equals negative π‘₯ plus one plus three 𝑖. Multiply through by negative one to get rid of this negative π‘₯-value. And we have a factor zero equals π‘₯ minus one minus three 𝑖. And our fourth factor is π‘₯ minus one minus three 𝑖. π‘˜ of π‘₯ equals negative five π‘₯ minus one times π‘₯ plus two times π‘₯ minus one plus three 𝑖 times π‘₯ minus one minus three 𝑖.

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