15. The polynomial $$ 2 x^{3}-3 a x^{2}+a x+b $$ has a factor $$ x-1 $$ and when divided by $$ x+2 $$, a remainder of -54 is obtained. Find the values of $$ a $$ and $$ b $$. With these values of $$ a $$ and $$ b $$, factorise the polynomial completely. Hence, or otherwise, solve $$ 2 x^{6}-9 x^{4}+3 x^{2}+4=0 $$. [Answer Key] $$ a=3, b=4 \quad x= \pm 1, \pm 2 $$

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To solve the problem, we will follow these steps:

1. **Use the Factor Theorem**: Since $$ x - 1 $$ is a factor of the polynomial $$ 2x^3 - 3ax^2 + ax + b $$, we can substitute $$ x = 1 $$ into the polynomial and set it equal to zero.

2. **Use the Remainder Theorem**: Since the remainder when divided by $$ x + 2 $$ is -54, we can substitute $$ x = -2 $$ into the polynomial and set it equal to -54.

3. **Set up the equations**: From the above two steps, we will derive two equations involving $$ a $$ and $$ b $$.

4. **Solve the system of equations**: We will solve the equations to find the values of $$ a $$ and $$ b $$.

5. **Factor the polynomial**: With the values of $$ a $$ and $$ b $$, we will factor the polynomial completely.

6. **Solve the equation $$ 2x^6 - 9x^4 + 3x^2 + 4 = 0 $$** using the factors obtained.

Let's start with the first two steps:

### Step 1: Substitute $$ x = 1 $$

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

This simplifies to:

$$
2 - 3a + a + b = 0 \implies 2 - 2a + b = 0 \implies b = 2a - 2 \quad \text{(Equation 1)}
$$

### Step 2: Substitute $$ x = -2 $$

$$
2(-2)^3 - 3a(-2)^2 + a(-2) + b = -54
$$

This simplifies to:

$$
2(-8) - 3a(4) - 2a + b = -54
$$
$$
-16 - 12a - 2a + b = -54 \implies b - 14a = -38 \implies b = 14a - 38 \quad \text{(Equation 2)}
$$

### Step 3: Set up the equations

Now we have two equations:

1. $$ b = 2a - 2 $$
2. $$ b = 14a - 38 $$

### Step 4: Solve the system of equations

Setting the two expressions for $$ b $$ equal to each other:

$$
2a - 2 = 14a - 38
$$

Rearranging gives:

$$
-2 + 38 = 14a - 2a \implies 36 = 12a \implies a = 3
$$

Now substituting $$ a = 3 $$ back into Equation 1 to find $$ b $$:

$$
b = 2(3) - 2 = 6 - 2 = 4
$$

Thus, we have:

$$
a = 3, \quad b = 4
$$

### Step 5: Factor the polynomial

Now substituting $$ a $$ and $$ b $$ back into the polynomial:

$$
2x^3 - 3(3)x^2 + 3x + 4 = 2x^3 - 9x^2 + 3x + 4
$$

We know $$ x - 1 $$ is a factor, so we can perform polynomial long division or synthetic division to factor it completely.

Let's perform synthetic division with $$ x = 1 $$:

$$
\begin{array}{r|rrrr}
1 & 2 & -9 & 3 & 4 \
  &   & 2 & -7 & -4 \
\hline
  & 2 & -7 & -4 & 0 \
\end{array}
$$

The result is $$ 2x^2 - 7x - 4 $$.

Now we need to factor $$ 2x^2 - 7x - 4 $$. We can use the quadratic formula:

$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{7 \pm \sqrt{(-7)^2 - 4(2)(-4)}}{2(2)} = \frac{7 \pm \sqrt{49 + 32}}{4} = \frac{7 \pm \sqrt{81}}{4} = \frac{7 \pm 9}{4}
$$

This gives us:

$$
x = \frac{16}{4} = 4 \quad \text{and} \quad x = \frac{-2}{4} = -\frac{1}{2}
$$

Thus, the complete factorization of the polynomial is:

$$
2x^3 - 9x^2 + 3x + 4 = (x - 1)(2x^2 - 7x - 4) = (x - 1)(x - 4)(2x + 1)
$$

### Step 6: Solve $$ 2x^6 - 9x^4 + 3x^2 + 4 = 0 $$

We can substitute $$ y = x^2 $$:

$$
2y^3 - 9y^2 + 3y + 4 = 0
$$

Using the factors we found, we can substitute back to find the roots:

1. $$ y = 1 $$ gives $$ x^2 = 1 $$ or $$ x = \pm 1 $$
2. $$ y = 4 $$ gives $$ x^2 = 4 $$ or $$ x = \pm 2 $$

Thus, the solutions to $$ 2x^6 - 9x^4 + 3x^2 + 4 = 0 $$ are:

$$
x = \pm 1, \pm 2
$$

### Final Answers

$$
a = 3, \quad b = 4 \quad \text{and} \quad x = \pm 1, \pm 2
$$
$$ a = 3 $$, $$ b = 4 $$, and the solutions to the equation are $$ x = \pm 1 $$ and $$ x = \pm 2 $$.

The polynomial has a factor and when divided by , a
remainder of -54 is obtained. Find the values of and . With these values of and ,
factorise the polynomial completely. Hence, or otherwise,
solve .
[Answer Key]

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The polynomial has a factor and when divided by , a remainder of -54 is obtained. Find the values of and . With these values of and , factorise the polynomial completely. Hence, or otherwise, solve . [Answer Key]