When the polynomial $$ 6 x^{4}+p x^{2}+q $$ is divided by $$ x^{2}-5 x+6 $$, the remainder is $$ 400 x-693 $$. Find the value of $$ p $$ and $$ q $$. [Answer Key] $$ p=2, q=3 $$

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We are given that
$$
6 x^4 + p x^2 + q
$$
divided by
$$
x^2 - 5x + 6
$$
leaves a remainder of
$$
400x - 693.
$$
Since the divisor is quadratic, the remainder must be of degree less than 2 (i.e. linear). We can express the division as
$$
6 x^4 + p x^2 + q = (x^2-5x+6) \cdot Q(x) + (400x-693),
$$
where $$Q(x)$$ is the quotient polynomial.

Because $$x^2-5x+6$$ factors as
$$
(x-2)(x-3)=0,
$$
its roots are $$x=2$$ and $$x=3$$. Substituting these roots into the equation will eliminate the $$Q(x)$$ term.

**Step 1. Substitute $$x=2$$**
$$
6(2^4) + p(2^2) + q = 400(2) - 693.
$$
Compute each term:
$$
6(16) = 96,\quad p(4)=4p,\quad 400(2)-693=800-693=107.
$$
Thus, we have:
$$
96 + 4p + q = 107.
$$
Subtracting 96 on both sides:
$$
4p + q = 11. \quad \text{(Equation 1)}
$$

**Step 2. Substitute $$x=3$$**
$$
6(3^4) + p(3^2) + q = 400(3) - 693.
$$
Compute each term:
$$
6(81) = 486,\quad p(9)=9p,\quad 400(3)-693=1200-693=507.
$$
Thus, we have:
$$
486 + 9p + q = 507.
$$
Subtracting 486 on both sides:
$$
9p + q = 21. \quad \text{(Equation 2)}
$$

**Step 3. Solve the system of equations**

Subtract Equation 1 from Equation 2:
$$
(9p+q)-(4p+q)=21-11,
$$
$$
5p=10,
$$
$$
p=2.
$$

Substitute $$p=2$$ into Equation 1:
$$
4(2)+q=11,
$$
$$
8+q=11,
$$
$$
q=3.
$$

Thus, the values are
$$
p=2,\quad q=3.
$$
$$ p = 2 $$ and $$ q = 3 $$.

When the polynomial is divided by , the remainder is
. Find the value of and .
[Answer Key]

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