Algebra Presentation on Topic Modulus Function and PolynomialsCreated by: Kanya and Albert Performance Task BINUS School Bekasi

1. Algebra on
Modulus Function
and Polynomials
By: Kanya & Albert
Start!

2. Solve and discuss
Polynomials
question
1st Objective 2nd Objective
Solve and discuss
Modulus Function
question
3rd Objective
Closing
MEETING OBJECTIVES
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3. Question
01
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4. The polynomial ax3+bx2+8x-2, where a and b are
constants, is denoted by p(x). It is given that
(2x+2) is a factor of p(x) and that when p(x) is
divided by (x-4) the remainder is 30
Find the values of a and b.
Question 1: Polynomials
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5. Step 1: Convert the vector to an (x=) form.
(2x+2) Remainder=0 (x-4) Remainder=30
x=-1 x=4
Step 2: Insert the converted factor to the equation
p(x)=ax3+bx2+8x-2
p(x)=ax3+bx2+8x-2
0=a(-1)3+b(-1)2+8(-1)-2
0=-a+b-8-2
0=-a+b-10
b=a+10
p(x)=ax3+bx2+8x-2
30=a(4)3+b(4)2+8(4)-2
30=64a+16b+30
0=64a+16b
Solution
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6. Solution
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Step 3: Substitute 64a+16b with b=a+10
=64a+16(a+10)
=64a+16a+160
80a=-160
a=-2
b=a+10
b=8
Step 4: Final Answers
a = -2 and b = 8
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7. Q.2
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8. Given that k is a positive constant,
solve the inequality I x - 9k I > I x - 18k I
Question 2: Modulus Function
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9. Step 1: Solve both sides of the equation and expand
(x-9k)2 > (x-18k)2
x2 -18xk + 81k2 > x2 -36xk + 324k2
Step 2: Cross out the same variable
x2 -18xk + 81k2 > x2 - 36xk + 324k2
-18xk + 81k2 > -36xk + 324k2
Step 3: Rearrange the equation
-18xk + 36xk + 81k2 - 324k2> 0
18xk - 243k2 > 0
Solution
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10. Step 4: Divide both sides with 9
18xk - 243k2 > 0
2xk - 27k2 > 0
Step 5: Solve x by putting constant k outside of the bracket
k(2x-27k) > 0
2x > 27k
x > 27k/2
x > 13.5k
Step 6: Final Answer
x > 13.5k
Solution
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11. THANK
YOU
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