14. If $$ x^{2}-5 x+6 $$ is a factor of $$ p(x)=x^{4}+a x^{3}-x^{2}+b x-72 $$ Find the values of $$ a $$ and $$ b $$. Then solve $$ p(x)=0 $$.

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UpStudy AI · Accepted Answer

To solve the problem, we start by determining the values of $$ a $$ and $$ b $$ such that $$ x^{2} - 5x + 6 $$ is a factor of the polynomial $$ p(x) = x^{4} + a x^{3} - x^{2} + b x - 72 $$.

### Step 1: Identify the Roots from the Given Factor

The quadratic $$ x^{2} - 5x + 6 $$ can be factored as:
$$
x^{2} - 5x + 6 = (x - 2)(x - 3)
$$
Thus, the roots are $$ x = 2 $$ and $$ x = 3 $$. These roots must satisfy $$ p(x) = 0 $$.

### Step 2: Formulate Equations Using the Roots

Substitute $$ x = 2 $$ and $$ x = 3 $$ into $$ p(x) $$:

1. For $$ x = 2 $$:
$$
p(2) = 2^{4} + a \cdot 2^{3} - 2^{2} + b \cdot 2 - 72 = 16 + 8a - 4 + 2b - 72 = 8a + 2b - 60 = 0
$$
Simplifying:
$$
8a + 2b = 60 \quad \text{(Equation 1)}
$$

2. For $$ x = 3 $$:
$$
p(3) = 3^{4} + a \cdot 3^{3} - 3^{2} + b \cdot 3 - 72 = 81 + 27a - 9 + 3b - 72 = 27a + 3b = 0
$$
Simplifying:
$$
27a + 3b = 0 \quad \text{(Equation 2)}
$$

### Step 3: Solve the System of Equations

From Equation 2:
$$
27a + 3b = 0 \implies 9a + b = 0 \implies b = -9a
$$

Substitute $$ b = -9a $$ into Equation 1:
$$
8a + 2(-9a) = 60 \implies 8a - 18a = 60 \implies -10a = 60 \implies a = -6
$$
Then:
$$
b = -9(-6) = 54
$$

### Step 4: Factor the Polynomial

Now, the polynomial becomes:
$$
p(x) = x^{4} - 6x^{3} - x^{2} + 54x - 72
$$
We know $$ x^{2} - 5x + 6 $$ is a factor, so:
$$
p(x) = (x^{2} - 5x + 6)(x^{2} + c x + d)
$$
Expanding and equating coefficients:
$$
x^{4} + (c - 5)x^{3} + (d - 5c + 6)x^{2} + (-5d + 6c)x + 6d = x^{4} - 6x^{3} - x^{2} + 54x - 72
$$
From this, we find:
$$
c - 5 = -6 \implies c = -1 \
d - 5(-1) + 6 = -1 \implies d + 11 = -1 \implies d = -12
$$
Thus:
$$
p(x) = (x^{2} - 5x + 6)(x^{2} - x - 12) = (x - 2)(x - 3)(x - 4)(x + 3)
$$

### Step 5: Solve $$ p(x) = 0 $$

The roots of the equation $$ p(x) = 0 $$ are:
$$
x = 2, \quad x = 3, \quad x = 4, \quad x = -3.
$$

### Final Answer

$$
\boxed{
\begin{aligned}
&a = -6, \
&b = 54, \
&\text{The solutions to } p(x) = 0 \text{ are } x = 2,\ 3,\ 4,\ \text{and } -3.
\end{aligned}
}
$$
$$ a = -6 $$, $$ b = 54 $$, and the solutions to $$ p(x) = 0 $$ are $$ x = 2 $$, $$ x = 3 $$, $$ x = 4 $$, and $$ x = -3 $$.

14. If \( x^{2}-5 x+6 \) is a factor of \( p(x)=x^{4}+a x^{3}-x^{2}+b x-72 \) Find the values of \( a \) and \( b \). Then solve \( p(x)=0 \).

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