Hello, Cathy!

I approached them differently . . . Here's #3.

$\displaystyle 3)\;\;x^2 +ax+b$ .is a factor of .$\displaystyle 2x^3+ax^2+13x+b$

Find the values of $\displaystyle a$ and $\displaystyle b$ I used long division . . . Code:

2x - a
---------------------------
x² + ax + b ) 2x³ + ax² + 13x + b
2x³ + 2ax² + 2bx
-----------------------
-ax² + (13-2b)x + b
-ax² - a²x - ab
--------------------
(13-2b+a²)x + (b+ab)

Since the division "comes out even", the remainder is zero.

Hence: .$\displaystyle \begin{array}{cccccccc}13-2b+a^2& = & 0& \Rightarrow & a^2+13 & = & 2b & {\color{blue}[1]}\\b+ab & = & 0 & \Rightarrow & b(a+1) & = & 0 & {\color{blue}[2]}\end{array}$

From [2], we have: .$\displaystyle b = 0$ .or .$\displaystyle a = -1$

If $\displaystyle b=0$, [1] becomes: .$\displaystyle a^2+13\:=\:0$ . . .which has no real roots

If $\displaystyle a=-1$, [1] becomes: .$\displaystyle 1+13 = 2b\quad\Rightarrow\quad b = 7 $

The problem is: .$\displaystyle (2x^3 - x^2 + 13x + 7) \div (x^2-x+7) \:=\:2x + 1$