AQA C1 2009 June — Question 4

Exam BoardAQA
ModuleC1 (Core Mathematics 1)
Year2009
SessionJune
TopicFactor & Remainder Theorem
TypeDirect remainder then factorise

4
  1. The polynomial \(\mathrm { p } ( x )\) is given by \(\mathrm { p } ( x ) = x ^ { 3 } - x + 6\).
    1. Find the remainder when \(\mathrm { p } ( x )\) is divided by \(x - 3\).
    2. Use the Factor Theorem to show that \(x + 2\) is a factor of \(\mathrm { p } ( x )\).
    3. Express \(\mathrm { p } ( x ) = x ^ { 3 } - x + 6\) in the form \(( x + 2 ) \left( x ^ { 2 } + b x + c \right)\), where \(b\) and \(c\) are integers.
    4. The equation \(\mathrm { p } ( x ) = 0\) has one root equal to - 2 . Show that the equation has no other real roots.
  2. The curve with equation \(y = x ^ { 3 } - x + 6\) is sketched below.
    \includegraphics[max width=\textwidth, alt={}, center]{5f1ff5fa-b6e8-4c4f-aef7-63eb947b299f-3_529_702_945_667} The curve cuts the \(x\)-axis at the point \(A ( - 2,0 )\) and the \(y\)-axis at the point \(B\).
    1. State the \(y\)-coordinate of the point \(B\).
    2. Find \(\int _ { - 2 } ^ { 0 } \left( x ^ { 3 } - x + 6 \right) \mathrm { d } x\).
    3. Hence find the area of the shaded region bounded by the curve \(y = x ^ { 3 } - x + 6\) and the line \(A B\).