Polynomial Theorems

Solved Examples on the Theorems of Polynomials

For ACT Students
The ACT is a timed exam...60 questions for 60 minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no negative penalty for a wrong answer.

For SAT Students
Any question labeled SAT-C is a question that allows a calculator.
Any question labeled SAT-NC is a question that does not allow a calculator.

For JAMB Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.
Unless specified otherwise, any question labeled JAMB is a question from JAMB Physics

For WASSCE Students: Unless specified otherwise:
Any question labeled WASCCE is a question from WASCCE Physics
Any question labeled WASSCE-FM is a question from the WASSCE Further Mathematics/Elective Mathematics

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Solve all questions.
Show all work.

(2.) ACT The function P(x) = x³ − 10x² + 9x − 90, where x is the number of items sold, models the profit P(x), in dollars, for Company A.
The company breaks even when the profit is $0.
How many items must Company A sell to break even?

$ A.\;\; 0 \\[3ex] B.\;\; 3 \\[3ex] C.\;\; 9 \\[3ex] D.\;\; 10 \\[3ex] E.\;\; 90 \\[3ex] $

$ P(x) = x^3 - 10x^2 + 9x - 90 \\[3ex] P(x) = 0 ...to\;\;break\;\;even \\[3ex] \implies \\[3ex] x^3 - 10x^2 + 9x - 90 = 0 \\[3ex] Factor\;\;by\;\;Grouping \\[3ex] x^2(x - 10) + 9(x - 10) = 0 \\[3ex] (x - 10)(x^2 + 9) = 0 \\[3ex] x - 10 = 0 \;\;\;OR\;\;\; x^2 + 9 = 0 \\[3ex] x = 10 \;\;\;OR\;\:\; x^2 = -9 \\[3ex] x = 10 \\[3ex] $ The other value of x is not a real number.
Company A must sell 10 items to break even.