Divide \( x^3 + 2x^2 - 5x - 6 \) by \( x-2 \) using algebraic long division.
\( x^2 + 4x + 3 \). (Remainder 0.)
Algebraic Methods
10 exam-style questions on algebraic division, factor/remainder theorem and partial fractions. Tap Show answer for full working.
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Given \( f(x) = x^3 - 4x^2 + x + 6 \), find \( f(2) \) and state whether \( (x-2) \) is a factor.
\( f(2)=8-16+2+6=0 \), so \( (x-2) \) is a factor.
Express \( \dfrac{5}{(x-1)(x+2)} \) in partial fractions.
\( \dfrac{5/3}{x-1} - \dfrac{5/3}{x+2} \) or \( \dfrac{5}{3}\left(\dfrac{1}{x-1}-\dfrac{1}{x+2}\right) \).
Express \( \dfrac{3x+1}{(x-1)(x+1)(x+2)} \) in partial fractions.
\( \dfrac{2/3}{x-1} + \dfrac{-1}{x+1} + \dfrac{1/3}{x+2} \).
The remainder when \( x^3 + ax^2 + bx + 6 \) is divided by \( x-1 \) is 4, and when divided by \( x+1 \) is 12. Find \( a \) and \( b \).
\( f(1)=1+a+b+6=4 \) ⇒ \( a+b=-3 \). \( f(-1)=-1+a-b+6=12 \) ⇒ \( a-b=7 \). Add: \( 2a=4 \) ⇒ \( a=2 \), \( b=-5 \).
Express \( \dfrac{x^2+1}{x(x-1)} \) in partial fractions.
Improper: \( 1 + \dfrac{1}{x(x-1)} \). Then \( \dfrac{1}{x(x-1)} = \dfrac{-1}{x} + \dfrac{1}{x-1} \). So \( 1 - \dfrac{1}{x} + \dfrac{1}{x-1} \).
Express \( \dfrac{2x+3}{(x+1)^2(x-1)} \) in partial fractions.
\( \dfrac{A}{x+1} + \dfrac{B}{(x+1)^2} + \dfrac{C}{x-1} \). \( A=-\dfrac{1}{4} \), \( B=-\dfrac{1}{2} \), \( C=\dfrac{5}{4} \).
Given \( f(x)=x^4 - 5x^2 + 4 \), factorise \( f(x) \) completely into linear factors.
\( f(x)=(x^2-1)(x^2-4)=(x-1)(x+1)(x-2)(x+2) \).
Express \( \dfrac{x^3+2}{x^2-1} \) in the form \( x + A + \dfrac{Bx+C}{x^2-1} \) and find \( A,B,C \).
Divide: \( x^3+2 = (x^2-1)(x) + x + 2 \). So \( x + \dfrac{x+2}{x^2-1} \). \( A=0 \), \( B=1 \), \( C=2 \).
Solve \( \dfrac{2}{x-1} - \dfrac{1}{x+2} = 1 \) using partial fractions or clearing denominators.
\( 2(x+2) - (x-1) = (x-1)(x+2) \) ⇒ \( 2x+4-x+1 = x^2+x-2 \) ⇒ \( x+5 = x^2+x-2 \) ⇒ \( x^2-7=0 \) ⇒ \( x=\pm\sqrt{7} \).