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Question 7: The polynomial P(x) = x³ − ax² + x − b is exactly divisible by x - 1 and leaves a remainder of -1 when divided by x - 2. (a) Find the values of a and b. (b) Factorise P(x) completely.

Mathematics

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6.

  1. Perform polynomial long division to find the constant term A.

    The denominator is (x+2)(x−1)=x2+x−2(x+2)(x-1) = x^2 + x - 2.
    Divide x2+3x−13x^2+3x-13 by x2+x−2x^2+x-2:

    x2+3x−13x2+x−2=1+(x2+3x−13)−(x2+x−2)x2+x−2\frac{x^2+3x-13}{x^2+x-2} = 1 + \frac{(x^2+3x-13) - (x^2+x-2)}{x^2+x-2} =1+2x−11(x+2)(x−1)= 1 + \frac{2x-11}{(x+2)(x-1)}

    So, A=1A=1.

  2. Decompose the remaining fraction into partial fractions.

    Let 2x−11(x+2)(x−1)=Bx+2+Cx−1\frac{2x-11}{(x+2)(x-1)} = \frac{B}{x+2} + \frac{C}{x-1}.
    Multiply both sides by (x+2)(x−1)(x+2)(x-1):

    2x−11=B(x−1)+C(x+2)2x-11 = B(x-1) + C(x+2)

  3. Solve for B and C.

    Set x=1x=1:

    2(1)−11=B(1−1)+C(1+2)2(1)-11 = B(1-1) + C(1+2) −9=0+3C-9 = 0 + 3C C=−3C = -3

    Set x=−2x=-2:

    2(−2)−11=B(−2−1)+C(−2+2)2(-2)-11 = B(-2-1) + C(-2+2) −4−11=−3B+0-4-11 = -3B + 0 −15=−3B-15 = -3B B=5B = 5

  4. Write the final expression.

    x2+3x−13(x+2)(x−1)=1+5x+2−3x−1\frac{x^2+3x-13}{(x+2)(x-1)} = \boxed{1 + \frac{5}{x+2} - \frac{3}{x-1}}

    where A=1A=1, B=5B=5, and C=−3C=-3.

    7.

    a)

  5. Use the Remainder Theorem to set up equations for aa and bb.

    Given P(x)=x3−ax2+x−bP(x) = x^3 - ax^2 + x - b.
    "exactly divisible by x−1x-1" means P(1)=0P(1) = 0.

    P(1)=(1)3−a(1)2+(1)−b=0P(1) = (1)^3 - a(1)^2 + (1) - b = 0 1−a+1−b=01 - a + 1 - b = 0 2−a−b=0  ⟹  a+b=2(1)2 - a - b = 0 \implies a+b=2 \quad (1)

    "leaves a remainder of −1-1 when divided by x−2x-2" means P(2)=−1P(2) = -1.

    P(2)=(2)3−a(2)2+(2)−b=−1P(2) = (2)^3 - a(2)^2 + (2) - b = -1 8−4a+2−b=−18 - 4a + 2 - b = -1 10−4a−b=−1  ⟹  4a+b=11(2)10 - 4a - b = -1 \implies 4a+b=11 \quad (2)

  6. Solve the system of linear equations for aa and bb.

    Subtract equation (1) from equation (2):

    (4a+b)−(a+b)=11−2(4a+b) - (a+b) = 11 - 2 3a=93a = 9 a=3a = 3

    Substitute a=3a=3 into equation (1):

    3+b=23+b=2 b=2−3b = 2-3 b=−1b = -1

    The values are a=3,b=−1\boxed{a=3, b=-1}.

    b)

  7. Write the polynomial with the found values of aa and bb.

    P(x)=x3−3x2+x−(−1)=x3−3x2+x+1P(x) = x^3 - 3x^2 + x - (-1) = x^3 - 3x^2 + x + 1.

  8. Factor out (x−1)(x-1) since P(1)=0P(1)=0.

    Use polynomial division or synthetic division:

    1 | 1  -3   1   1
  |    1  -2  -1
  ----------------
    1  -2  -1   0

    So, P(x)=(x−1)(x2−2x−1)P(x) = (x-1)(x^2 - 2x - 1).

  9. Factor the quadratic term x2−2x−1x^2 - 2x - 1.

    Use the quadratic formula x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2-4ac}}{2a} for x2−2x−1=0x^2 - 2x - 1 = 0:

    x=−(−2)±(−2)2−4(1)(−1)2(1)x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2(1)} x=2±4+42x = \frac{2 \pm \sqrt{4 + 4}}{2} x=2±82x = \frac{2 \pm \sqrt{8}}{2} x=2±222x = \frac{2 \pm 2\sqrt{2}}{2} x=1±2x = 1 \pm \sqrt{2}

    So the factors are (x−(1+2))(x - (1+\sqrt{2})) and (x−(1−2))(x - (1-\sqrt{2})).
    Therefore, P(x)=(x−1)(x−1−2)(x−1+2)P(x) = \boxed{(x-1)(x - 1 - \sqrt{2})(x - 1 + \sqrt{2})}.

    8.
    Mr Moyo's house (A) is at (−5,3)(-5, 3).
    Well (P) is at (2,1)(2, 1).
    P is the midpoint of the line segment AD, where D is Mr Dube's house.

    a)

  10. Use the midpoint formula to find the coordinates of Mr Dube's house.

    Let Dube's house be (xD,yD)(x_D, y_D). The midpoint formula is P=(xA+xD2,yA+yD2)P = \left(\frac{x_A+x_D}{2}, \frac{y_A+y_D}{2}\right).

    (2,1)=(−5+xD2,3+yD2)(2, 1) = \left(\frac{-5+x_D}{2}, \frac{3+y_D}{2}\right)

    Equating the x-coordinates:

    2=−5+xD22 = \frac{-5+x_D}{2} 4=−5+xD4 = -5+x_D xD=9x_D = 9

    Equating the y-coordinates:

    1=3+yD21 = \frac{3+y_D}{2} 2=3+yD2 = 3+y_D yD=−1y_D = -1

    The coordinates of Mr Dube's house are (9,−1)\boxed{(9, -1)}.

    b)

  11. Use the distance formula to find the distance between Mr Moyo's house and Mr Dube's house.

    Distance AD=(xD−xA)2+(yD−yA)2AD = \sqrt{(x_D-x_A)^2 + (y_D-y_A)^2}.
    A=(−5,3)A = (-5, 3) and D=(9,−1)D = (9, -1).

    AD=(9−(−5))2+(−1−3)2AD = \sqrt{(9 - (-5))^2 + (-1 - 3)^2} AD=(14)2+(−4)2AD = \sqrt{(14)^2 + (-4)^2} AD=196+16AD = \sqrt{196 + 16} AD=212AD = \sqrt{212}

    Simplify the radical: 212=4×53212 = 4 \times 53.

    AD=4×53=253AD = \sqrt{4 \times 53} = 2\sqrt{53}

    The distance from Mr Moyo to Mr Dube's house is 253 units\boxed{2\sqrt{53} \text{ units}}.

    c)

  12. Identify the geometric meaning of the line of points.

    A line of points equidistant from two given points is the perpendicular bisector of the line segment connecting those two points.

  13. Find the gradient of the line segment AD.

    A=(−5,3)A = (-5, 3) and D=(9,−1)D = (9, -1).
    Gradient mAD=yD−yAxD−xA=−1−39−(−5)=−414=−27m_{AD} = \frac{y_D - y_A}{x_D - x_A} = \frac{-1 - 3}{9 - (-5)} = \frac{-4}{14} = -\frac{2}{7}.

  14. Find the gradient of the perpendicular bisector.

    The gradient of the perpendicular bisector m⊥m_{\perp} is the negative reciprocal of mADm_{AD}.

    m⊥=−1mAD=−1−2/7=72m_{\perp} = -\frac{1}{m_{AD}} = -\frac{1}{-2/7} = \frac{7}{2}

  15. Use the midpoint P and the perpendicular gradient to find the equation of the line.

    The perpendicular bisector passes through the midpoint P (2,1)(2, 1).
    Using the point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1):

    y−1=72(x−2)y - 1 = \frac{7}{2}(x - 2)

    Multiply by 2 to clear the fraction:

    2(y−1)=7(x−2)2(y - 1) = 7(x - 2) 2y−2=7x−142y - 2 = 7x - 14

    Rearrange into the general form Ax+By+C=0Ax+By+C=0:

    7x−2y−14+2=07x - 2y - 14 + 2 = 0 7x−2y−12=0\boxed{7x - 2y - 12 = 0}
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