PreCalc 12 Help!!?

Let f(x) = 5x² + bx + 14

When f(x) is divided by x - 3, the remainder is 158.
By Remainder Theorem, f(3) = 158
5(3)² + b(3) + 14 = 158
b = 33
Hence, f(x) = 5x² + 33x + 14

When f(x) is divided by x + 5, the remainder
= f(-5)
= 5(-5)² + 33(-5) + 14
= -26

Do you mean 5x² + bx + 14?

Let p(x) = 5x² + bx + 14.

When dividing p(x) by (x - 3), the remainder is 158.

p(3) = 158
5(3)² + b(3) + 14 = 158
b = 33

Dividing by (x + 5), the remainder is p(-5).

p(-5) = 5(-5)² + 33(-5) + 14 = -26

5x² + bx + 14

If that is divided by (x - 3) the remainder is 158.  That means that the expression is equal to 158 when x = 3.  We can use this information to solve for the unknown "b":

5(3)² + b(3) + 14 = 158
5(9) + 3b + 14 = 158
45 + 3b + 14 = 158
3b + 59 = 158
3b = 99
b = 33

Now that we know what "b" is we have this expression:

5x² + 33x + 14

What's the remainder when divided by (x + 5)?  Simplify when x = -5 and we have our answer:

5(-5)² + 33(-5) + 14
5(25) - 165 + 14
125 - 165 + 14
-26