# Finding Values of A and B

• Nov 17th 2009, 10:31 AM
KingV15
Finding Values of A and B
Determine a and b so that (x-1)^2 divides evenly into the polynomial
f(x)=ax^4+bx^3+1.
First you must divide x-1 into f(x), the divide x-1 again into quotient.
• Nov 17th 2009, 08:40 PM
I4talent
Quote:

Originally Posted by KingV15
Determine a and b so that (x-1)^2 divides evenly into the polynomial
f(x)=ax^4+bx^3+1.
First you must divide x-1 into f(x), the divide x-1 again into quotient.

Let \$\displaystyle ax^4+bx^3+1 = (x-1)(x-1)(cx^2+dx+e)
\$

\$\displaystyle
ax^4+bx^3+1 = (x-1)(x-1)(cx^2+dx+e)\$

\$\displaystyle ax^4+bx^3+1 = (x^2-2x+1)(cx^2+dx+e)\$

\$\displaystyle ax^4+bx^3+1 = x^2(cx^2+dx+e)-2x(cx^2+dx+e)+1(cx^2+dx+e)\$

\$\displaystyle ax^4+bx^3+1 = cx^4+dx^3+ex^2-2cx^3-2dx^2-2ex+cx^2+dx+e\$

\$\displaystyle ax^4+bx^3+1 = cx^4+dx^3-2cx^3+ex^2-2dx^2cx^2-2ex+dx+e\$

\$\displaystyle ax^4+bx^3+1 = (c)x^4+(d-2c)x^3+(e-2d+c)x^2+(-2e+d)x+e\$

Comparing the coefficients:

\$\displaystyle \boxed{e = 1}. \$

\$\displaystyle -2e+d = 0 \Rightarrow -2(1)+d = 0 \Rightarrow -2+d = 0 \Rightarrow \boxed{d = 2}\$

\$\displaystyle e-2d+c = 0 \Rightarrow 1-2(2)+c = 0 \Rightarrow -3+c = 0 \Rightarrow \boxed{c = 3}\$

\$\displaystyle d-2c = b \Rightarrow 2-2(3) = b \Rightarrow 2-6 = b \Rightarrow \boxed{-4 = b}\$

\$\displaystyle a = c \Rightarrow \boxed{a = 3}\$

Hence \$\displaystyle \boxed{a = 3}\$ and \$\displaystyle \boxed{b = -4}\$