f(x)=(x-k)q(x)+ r has to pass a certain point

Question

I am confused at this question.

Here is the question: Use the form f(x)=(x-k)q(x)+ r to create a cubic function that passes through the point (-3,1) and falls to the right.

I've never seen the form f(x)=(x-k)q(x)+ r

Please Show work so I understand the solution better

Answer 1

(x - k) is a binomial expression. You're used to them when you factor a quadratic like x² - 2x - 8 into (x - 4)(x + 2) where one of the "(x - k)'s is (x - 4) with k = 4.

In this case, since they are asking for a cubic function, then q(x) would need to be a quadratic because only a quadratic multiplied by a linear factor like (x - k) will yield a cubic.

Then, they want the product of (x - k) and q(x) to require an additional "remainder" "r", in other words, when you divide f(x) by (x - k), you get q(x) + remainder "r".

Since they tell you the cubic "falls right" that means it must have a leading coefficient that is negative. Since (x - k) can't help with that, we need q(x) to be a quadratic with a leading coefficient that is negative.

f(x) = (x - k)(ax² + bx + c) + r where a is negative. Lets arbitrarily pick k = 1 (for no particular reason) and let's pick a = -2 (also no particular reason)

So, that will give us f(x) = (x - 1)(-2x² + bx + c) + r

Plugging in the point (-3, 1) we get:

1 = (-3 - 1)(-2(-3)² + b(-3) + c) + r

1 = (-4)(-18 -3b + c) + r

There are too many variables, so let's arbitrarily pick b = -6 and c = 1, That gives us:

1 = (-4)(-18 +18 + 1) + r

Simplifying, we get r = 5

That gives us f(x) = (x - 1)(-2x² - 6x + 1) + 5

Of course there are lots of possibilities for this. You can pick your own values of 3 of the constants k, a, b, c, and r and solve for remaining one, just make sure a is a negative number.