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Similar to Algebra Presentation on Topic Modulus Function and Polynomials
Algebra Presentation on Topic Modulus Function and Polynomials
- 1. Algebra on Modulus Function andPolynomials By: Kanya & Albert Start!
- 2. Solve and discuss Polynomials question 1stObjective 2nd Objective Solve and discuss Modulus Function question 3rd Objective Closing MEETING OBJECTIVES Back Agenda Next
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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 Back Agenda Next
- 5. Step 1: Convertthe 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 Agenda Next
- 6. Solution Agenda Next 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 Back
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- 8. Given that kis a positive constant, solve the inequality I x - 9k I > I x - 18k I Question 2: Modulus Function Back Agenda Next
- 9. Step 1: Solveboth 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 Agenda Next
- 10. Step 4: Divideboth 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 Agenda Next
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