Math Problem Statement

2x qubed plus ax squared plus bx plus 1 is divisible by 2x minus 1 and x plus 1 find a and b

Solution

We are given that the expression 2x3+ax2+bx+12x^3 + ax^2 + bx + 1 is divisible by (2x1)(x+1)(2x - 1)(x + 1). Our goal is to find the values of aa and bb.

Step 1: Division setup

We are told that the polynomial is divisible by 2x12x - 1 and x+1x + 1. This means both 2x12x - 1 and x+1x + 1 are factors of the cubic polynomial 2x3+ax2+bx+12x^3 + ax^2 + bx + 1.

Step 2: Use of Remainder Theorem

The Remainder Theorem tells us that if a polynomial f(x)f(x) is divisible by (xr)(x - r), then f(r)=0f(r) = 0. We can use this on the two roots of the factors 2x12x - 1 and x+1x + 1.

Root of 2x1=02x - 1 = 0

From 2x1=02x - 1 = 0, we find x=12x = \frac{1}{2}. Substituting x=12x = \frac{1}{2} into the cubic polynomial:

2(12)3+a(12)2+b(12)+1=02\left(\frac{1}{2}\right)^3 + a\left(\frac{1}{2}\right)^2 + b\left(\frac{1}{2}\right) + 1 = 0

This simplifies to:

2(18)+a(14)+b(12)+1=02\left(\frac{1}{8}\right) + a\left(\frac{1}{4}\right) + b\left(\frac{1}{2}\right) + 1 = 0

14+a4+b2+1=0\frac{1}{4} + \frac{a}{4} + \frac{b}{2} + 1 = 0

Multiply through by 4 to clear the denominators:

1+a+2b+4=01 + a + 2b + 4 = 0

This simplifies to:

a+2b=5(Equation 1)a + 2b = -5 \quad \text{(Equation 1)}

Root of x+1=0x + 1 = 0

From x+1=0x + 1 = 0, we find x=1x = -1. Substituting x=1x = -1 into the cubic polynomial:

2(1)3+a(1)2+b(1)+1=02(-1)^3 + a(-1)^2 + b(-1) + 1 = 0

This simplifies to:

2(1)+a(1)+b(1)+1=02(-1) + a(1) + b(-1) + 1 = 0

2+ab+1=0-2 + a - b + 1 = 0

This simplifies to:

ab=1(Equation 2)a - b = 1 \quad \text{(Equation 2)}

Step 3: Solve the system of equations

We now have the following system of equations:

  1. a+2b=5a + 2b = -5
  2. ab=1a - b = 1

Solve for aa and bb

From Equation 2, we can express aa in terms of bb:

a=1+ba = 1 + b

Substitute this into Equation 1:

(1+b)+2b=5(1 + b) + 2b = -5

Simplify:

1+3b=51 + 3b = -5

3b=63b = -6

b=2b = -2

Now substitute b=2b = -2 back into the expression for aa:

a=1+(2)=1a = 1 + (-2) = -1

Final Answer:

The values of aa and bb are: a=1,b=2a = -1, \quad b = -2

Would you like further details on this solution or any clarifications?


Here are 5 related questions to explore further:

  1. How do you factor polynomials using the Remainder Theorem?
  2. What are other methods for solving cubic polynomials?
  3. How does synthetic division help in solving polynomial division?
  4. What is the role of the Factor Theorem in polynomial division?
  5. How can you check if a polynomial is divisible by a quadratic factor?

Tip: When solving polynomial equations, try to factorize them first by identifying potential roots, which can simplify the process significantly.

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Math Problem Analysis

Mathematical Concepts

Algebra
Polynomial Division
Cubic Polynomials

Formulas

Cubic Polynomial: 2x^3 + ax^2 + bx + 1
Factorization: (2x - 1)(x + 1)
System of Linear Equations

Theorems

Remainder Theorem
Factor Theorem

Suitable Grade Level

Grades 10-12