Activity 2: Learning Remainder Theorem
1. Make 2 paper cuboids.
2. Write the polynomials in a single variable with degree 1 to 5 on 5 faces of one of a cuboid.
Write a linear polynomials on 4 faces of another cuboid in the same variable.
3. Play the two cuboids to get the combination of a polynomial p(x) and a linear polynomial
g(x).
4. Find the zero of linear polynomial by equating it to 0. Find the value of polynomial p(x)
for zero of the linear polynomial
5. Divide the polynomial by the linear polynomial by long division method and find its
remainder
6. Cross check the results for step 5 with that of step6. Is the remainder same by both
methods?
7. Write down the combination of a polynomial and a linear polynomial, where a linear
polynomial is a factor of polynomial. Justify your answer.
Answers
Answer:
Step 6: The remainder is the same by both methods.
Explanation:
First, take any polynomials for 2 cuboids as per requirements given in Step 2.
1 Example for Cuboid 1 : x^2 + 4 ( degree= 2)
1 Example for Cuboid 2 : 3x+ 8 ( all polynomials should have degree =1)
Then take any random polynomial out of the selected ones for both cuboids. Let the polynomial for cuboid 1 be p(x) and polynomial for cuboid 2 g(x).
Step 4: Find the zero of the linear polynomial, that is the polynomial you took for the second cuboid in Step 2.
Example: If the linear polynomial is taken as 3x + 8
3x + 8 = 0
3x= (-8)
x= -8/3 ( To avoid getting fractional answers, take easy linear polynomials)
Use the answer that you get in the above step in p(x) by taking "x" as the answer you got in the above step.
So if your p(x) is x^2+4 and x=2 then your equation will be,
x^2 +4= 2^2 + 4
= 4+4 = 8
This was using the Remainder Theorem. Now, we solve the same thing using the long division method.
Step 5: Divide the initially taken polynomial by the initially taken polynomial using the long division method.
Step 6 : You will observe the answers in both methods are the same.
Thus , Verified.
Step 7: Give an example of a polynomial that has a linear polynomial as a factor.