Question Video: Finding the Set of Zeros of a Function Mathematics • 10th Grade

Find the set of zeros of the function 𝑓(π‘₯) = (π‘₯Β² βˆ’ 2π‘₯ βˆ’ 15)(π‘₯Β² βˆ’ 14π‘₯ + 45).

04:13

Video Transcript

Find the set of zeros of the function 𝑓 of π‘₯ equals π‘₯ squared minus two π‘₯ minus 15 multiplied by π‘₯ squared minus 14π‘₯ plus 45.

We’re asked to find the set of zeros of a given function 𝑓 of π‘₯. 𝑓 of π‘₯ is the product of two quadratic factors. But if we distributed the parentheses, we’d see that it is a polynomial. The zeros, or roots, of a polynomial 𝑓 of π‘₯ are the values π‘₯ equals π‘Ž such that 𝑓 of π‘Ž is equal to zero. Essentially, they are the values of the variable that make the value of the function zero. To find these values of π‘₯, we set the expression for 𝑓 of π‘₯ equal to zero.

Now, as we’ve already said, this function 𝑓 of π‘₯ is the product of two quadratic functions. If the product of these functions is to be equal to zero, it follows that at least one of the quadratic factors themselves must be equal to zero. Hence, either π‘₯ squared minus two π‘₯ minus 15 must be equal to zero or π‘₯ squared minus 14π‘₯ plus 45 must be equal to zero. We now need to solve each of these quadratic equations. And there are a variety of methods we could use, such as using the quadratic formula or completing the square. However, if a quadratic equation can be solved by factoring, this is usually the most efficient way, so let’s try this first.

Considering the quadratic π‘₯ squared minus two π‘₯ minus 15 first, the coefficient of π‘₯ squared is one. So, the first term in each linear factor will simply be π‘₯. To complete the two factors, we then look for two numbers whose sum is the coefficient of π‘₯, which is negative two, and whose product is equal to the constant term of negative 15. Those two numbers are negative five and positive three. So, the factored form of the first quadratic equation is π‘₯ minus five multiplied by π‘₯ plus three equals zero.

We now have the product of two linear factors equal to zero. And once again, it follows that if the product is equal to zero, at least one of the individual factors must be equal to zero. Hence, we have the two linear equations π‘₯ minus five equals zero and π‘₯ plus three equals zero. These equations can be solved in one step, by adding five to both sides for the first equation and subtracting three from both sides of the second equation, to give the solutions π‘₯ equals five and π‘₯ equals negative three. Two of the zeros of the function 𝑓 of π‘₯ are therefore negative three and five.

We then follow the same process to factor the second quadratic. This time, we’re looking for two numbers with a sum of negative 14 and a product of 45. Those two numbers are negative five and negative nine. So, the factored form of the second quadratic equation is π‘₯ minus five multiplied by π‘₯ minus nine equals zero. We then set each linear factor equal to zero and solve the resulting equations to give π‘₯ equals five or π‘₯ equals nine.

Notice that one of these values, π‘₯ equals five, has already been identified as a zero of 𝑓 of π‘₯ using the first quadratic factor. We don’t need to include this value twice when listing the zeros. So, adding just the value nine to our list of zeros, we have that the set of zeros of the function 𝑓 of π‘₯, which are all the values of π‘₯ such that 𝑓 of π‘₯ is equal to zero, is the set containing the values negative three, five, and nine.

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