Video Transcript
Find, by factoring, the zeros of
the function π of π₯ equals nine π₯ squared plus nine π₯ minus 40.
To find the zeros of a function, we
set the function equal to zero, which means that nine π₯ squared plus nine π₯ minus
40 equals zero. We can solve this equation for π₯
to find the zeros of π.
The question asks us to find the
zeros by factoring. But here we have the slight
complication that the polynomial is nonmonic. So we will need to also choose the
correct coefficients for the π₯-terms in the factors. To factorize a nonmonic quadratic,
we can use the grouping method. We need to find a factor pair of
ππ that adds to give π, where π is the coefficient of π₯ squared, π is the
coefficient of π₯, and π is the constant coefficient. Taking the product, ππ is nine
times negative 40, which is negative 360. π is of course just nine. So, we need to find a pair of
numbers that will multiply to give negative 360 and add to give nine.
Since the numbers multiply to give
a negative number, one of them must be negative and the other must be positive. The first pair of numbers we might
think of that adds to make nine are negative one and 10. But clearly these numbers are too
small in magnitude to multiply to negative 360. To find any other pair of numbers
that will add to give nine, we can subtract one from the negative number and add one
to the positive number. We can skip a few to, say, negative
11 and 20. These multiply to give negative
220. So weβre still a little short. If we proceed in this way, we
eventually find that negative 15 and positive 24 add together to make nine and
multiply together to make negative 360.
A perhaps faster way to do this is
by listing all of the factor pairs of 360, picking one of the pair to be negative
and seeing which pair add to make nine. In this case, however, 360 happens
to be a highly divisible number. So this might actually take
longer. Now that we have the factor pair,
we can rewrite the original nine π₯ in the equation as negative 15π₯ plus 24π₯. The grouping method ensures that we
now have the same common factor between the first two terms and the last two terms,
in this case three π₯ minus five. We can rewrite the expression on
the left-hand side as three π₯ times three π₯ minus five plus eight times three π₯
minus five. This same factor, three π₯ minus
five, is also common to both of these terms. So we can rewrite this again as
three π₯ plus eight times three π₯ minus five.
We now have a product of two
binomials that are linear in π₯ and have no more common factors. So we cannot factorize any
further. We have a product of two terms that
is equal to zero. Therefore, at least one of the
terms must be equal to zero itself. This means that either three π₯
plus eight is equal to zero or three π₯ minus five is equal to zero. We can solve for π₯ in the first
equation by subtracting eight and dividing by three, giving π₯ equals negative eight
over three. And we can solve for π₯ in the
second equation by adding five and dividing by three, giving π₯ equals five over
three. Therefore, the set of zeros of the
function π is negative eight over three and five over three.
We can verify that these are in
fact the zeros of π by substituting the values into the original polynomial. π of negative eight over three is
equal to nine times negative eight over three all squared plus nine times negative
eight over three minus 40, which is equal to nine times 64 over nine plus negative
72 over three minus 40, which is equal to 64 minus 24 minus 40, which is indeed
equal to zero. And likewise, we can take π of
five over three, which eventually simplifies to 25 plus 15 minus 40, which is also
equal to zero. Therefore, these are indeed the
zeros of π. And since π is a quadratic
function, we know that there cannot be any more zeros.
