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

We are given that the expression \( 2x^3 + ax^2 + bx + 1 \) is divisible by \( (2x - 1)(x + 1) \). Our goal is to find the values of \( a \) and \( b \).

### Step 1: Division setup
We are told that the polynomial is divisible by \( 2x - 1 \) and \( x + 1 \). This means both \( 2x - 1 \) and \( x + 1 \) are factors of the cubic polynomial \( 2x^3 + ax^2 + bx + 1 \).

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

#### Root of \( 2x - 1 = 0 \)
From \( 2x - 1 = 0 \), we find \( x = \frac{1}{2} \). Substituting \( x = \frac{1}{2} \) into the cubic polynomial:

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

This simplifies to:

\[
2\left(\frac{1}{8}\right) + a\left(\frac{1}{4}\right) + b\left(\frac{1}{2}\right) + 1 = 0
\]

\[
\frac{1}{4} + \frac{a}{4} + \frac{b}{2} + 1 = 0
\]

Multiply through by 4 to clear the denominators:

\[
1 + a + 2b + 4 = 0
\]

This simplifies to:

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

#### Root of \( x + 1 = 0 \)
From \( x + 1 = 0 \), we find \( x = -1 \). Substituting \( x = -1 \) into the cubic polynomial:

\[
2(-1)^3 + a(-1)^2 + b(-1) + 1 = 0
\]

This simplifies to:

\[
2(-1) + a(1) + b(-1) + 1 = 0
\]

\[
-2 + a - b + 1 = 0
\]

This simplifies to:

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

### Step 3: Solve the system of equations
We now have the following system of equations:
1. \( a + 2b = -5 \)
2. \( a - b = 1 \)

#### Solve for \( a \) and \( b \)
From Equation 2, we can express \( a \) in terms of \( b \):

\[
a = 1 + b
\]

Substitute this into Equation 1:

\[
(1 + b) + 2b = -5
\]

Simplify:

\[
1 + 3b = -5
\]

\[
3b = -6
\]

\[
b = -2
\]

Now substitute \( b = -2 \) back into the expression for \( a \):

\[
a = 1 + (-2) = -1
\]

### Final Answer:
The values of \( a \) and \( b \) are:
\[
a = -1, \quad b = -2
\]

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

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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.

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

This solution explains how to find the values of a and b for the cubic polynomial 2x^3 + ax^2 + bx + 1, which is divisible by (2x - 1)(x + 1). Using the Remainder and Factor Theorems, it involves solving a system of equations. Suitable for Grades 10-12.

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