Powerpoint presentation by: Jobry and Deno HS 12A
Problems on Polynomials
Given that x-1 is the divisor of x2+2+1 with remainder “b”. find the
solution of bx2+4bx-12b where it is also a factor to 3x2-5x-2.
Problems on Polynomials
1. Find the value of “b”
by using long division
Solution
2. Input the value of b and use it
to solve the first equation
3. Check for which value of X gives a result of 0 when
you input it to the second equation. When X=0, X is a
factor to the equation
We can conclude from the following working, 2 is the solution to the problem.
Problems on Modulus
Function
Problems on Modulus Function
Solve the inequality |x-7√3|<|x+10√3|
Solution
1. Since
We square both sides and get the
expanded equations.
2. We rearrange the equations. Move the
x and the x2 to the left side and 147 to the
right side. The x2 will be eliminated.
3. Solve the inequality. Be
careful, we have to reverse the
inequality sign, because we
divided both sides by negative
value. Simplify the root sign by
multiplying both numerator
and denominator with its root
value.
Algebra Presentation on Topic Modulus Function and Polynomials

Algebra Presentation on Topic Modulus Function and Polynomials

  • 1.
    Powerpoint presentation by:Jobry and Deno HS 12A
  • 2.
  • 3.
    Given that x-1is the divisor of x2+2+1 with remainder “b”. find the solution of bx2+4bx-12b where it is also a factor to 3x2-5x-2. Problems on Polynomials
  • 4.
    1. Find thevalue of “b” by using long division Solution 2. Input the value of b and use it to solve the first equation 3. Check for which value of X gives a result of 0 when you input it to the second equation. When X=0, X is a factor to the equation We can conclude from the following working, 2 is the solution to the problem.
  • 5.
  • 6.
    Problems on ModulusFunction Solve the inequality |x-7√3|<|x+10√3|
  • 7.
    Solution 1. Since We squareboth sides and get the expanded equations. 2. We rearrange the equations. Move the x and the x2 to the left side and 147 to the right side. The x2 will be eliminated. 3. Solve the inequality. Be careful, we have to reverse the inequality sign, because we divided both sides by negative value. Simplify the root sign by multiplying both numerator and denominator with its root value.