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Given the polynomial g(x) = 3x³ - 7x² + 3x - 5x - 1, find the remainder and quotient when g(x) is divided by (x² - 2x - 1)

Mathematics

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Graph of g(x) = 3x³ − 7x² + 3x − 5x − 1

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Key points of g(x) = 3x³ − 7x² + 3x − 5x − 1
Pointxy
Maximum−0.13−0.86
y-intercept0−1
Minimum1.69−9.89
x-intercept2.630

Step-by-step solution

41 steps

QUESTION FOUR

a) Leaving your answer in surd form find the value of:

(i) sin⁡15∘\sin 15^\circ

  1. Use the angle subtraction formula for sine

    sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin(A-B) = \sin A \cos B - \cos A \sin B.
    Let A=45∘A = 45^\circ and B=30∘B = 30^\circ.

    sin⁡15∘=sin⁡(45∘−30∘)=sin⁡45∘cos⁡30∘−cos⁡45∘sin⁡30∘\sin 15^\circ = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ

  2. Substitute the known exact values of the trigonometric functions.

    sin⁡15∘=(22)(32)−(22)(12)\sin 15^\circ = \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right) sin⁡15∘=64−24\sin 15^\circ = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}

  3. Combine the terms.

    sin⁡15∘=6−24\boxed{\sin 15^\circ = \frac{\sqrt{6} - \sqrt{2}}{4}}

    (ii) tan⁡105∘\tan 105^\circ

  4. Use the angle addition formula for tangent

    tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}.
    Let A=60∘A = 60^\circ and B=45∘B = 45^\circ.

    tan⁡105∘=tan⁡(60∘+45∘)=tan⁡60∘+tan⁡45∘1−tan⁡60∘tan⁡45∘\tan 105^\circ = \tan(60^\circ + 45^\circ) = \frac{\tan 60^\circ + \tan 45^\circ}{1 - \tan 60^\circ \tan 45^\circ}

  5. Substitute the known exact values of the trigonometric functions.

    tan⁡105∘=3+11−3⋅1\tan 105^\circ = \frac{\sqrt{3} + 1}{1 - \sqrt{3} \cdot 1} tan⁡105∘=3+11−3\tan 105^\circ = \frac{\sqrt{3} + 1}{1 - \sqrt{3}}

  6. Rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, (1+3)(1+\sqrt{3}).

    tan⁡105∘=(3+1)(1+3)(1−3)(1+3)\tan 105^\circ = \frac{(\sqrt{3} + 1)(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})} tan⁡105∘=3+3+1+31−3\tan 105^\circ = \frac{\sqrt{3} + 3 + 1 + \sqrt{3}}{1 - 3} tan⁡105∘=4+23−2\tan 105^\circ = \frac{4 + 2\sqrt{3}}{-2}

  7. Simplify the expression.

    tan⁡105∘=−2−3\boxed{\tan 105^\circ = -2 - \sqrt{3}}

    b) Prove the following identities:

    (i) 11−sin⁡x−11+sin⁡x=2tan⁡xsec⁡x\frac{1}{1-\sin x} - \frac{1}{1+\sin x} = 2 \tan x \sec x

  8. Start with the Left Hand Side (LHS) and find a common denominator.

    LHS=11−sin⁡x−11+sin⁡x\text{LHS} = \frac{1}{1-\sin x} - \frac{1}{1+\sin x} LHS=(1+sin⁡x)−(1−sin⁡x)(1−sin⁡x)(1+sin⁡x)\text{LHS} = \frac{(1+\sin x) - (1-\sin x)}{(1-\sin x)(1+\sin x)}

  9. Simplify the numerator and use the identity (1−sin⁡x)(1+sin⁡x)=1−sin⁡2x=cos⁡2x(1-\sin x)(1+\sin x) = 1 - \sin^2 x = \cos^2 x in the denominator.

    LHS=1+sin⁡x−1+sin⁡x1−sin⁡2x\text{LHS} = \frac{1+\sin x - 1 + \sin x}{1 - \sin^2 x} LHS=2sin⁡xcos⁡2x\text{LHS} = \frac{2\sin x}{\cos^2 x}

  10. Rewrite the expression in terms of tan⁡x\tan x and sec⁡x\sec x using tan⁡x=sin⁡xcos⁡x\tan x = \frac{\sin x}{\cos x} and sec⁡x=1cos⁡x\sec x = \frac{1}{\cos x}.

    LHS=2(sin⁡xcos⁡x)(1cos⁡x)\text{LHS} = 2 \left(\frac{\sin x}{\cos x}\right) \left(\frac{1}{\cos x}\right) LHS=2tan⁡xsec⁡x\text{LHS} = 2 \tan x \sec x

    This is equal to the Right Hand Side (RHS).

    LHS=RHS\text{LHS} = \text{RHS}

    Therefore, the identity is proven.

    (ii) tan⁡x=sin⁡2x+sin⁡x1+cos⁡2x+cos⁡x\tan x = \frac{\sin 2x + \sin x}{1+\cos 2x + \cos x}

  11. Start with the Right Hand Side (RHS) and use the double angle identities

    sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x \cos x and cos⁡2x=2cos⁡2x−1\cos 2x = 2\cos^2 x - 1.

    RHS=sin⁡2x+sin⁡x1+cos⁡2x+cos⁡x\text{RHS} = \frac{\sin 2x + \sin x}{1+\cos 2x + \cos x} RHS=2sin⁡xcos⁡x+sin⁡x1+(2cos⁡2x−1)+cos⁡x\text{RHS} = \frac{2\sin x \cos x + \sin x}{1 + (2\cos^2 x - 1) + \cos x}

  12. Simplify the numerator and denominator.

    RHS=2sin⁡xcos⁡x+sin⁡x2cos⁡2x+cos⁡x\text{RHS} = \frac{2\sin x \cos x + \sin x}{2\cos^2 x + \cos x}

  13. Factor out common terms from the numerator and denominator.

    RHS=sin⁡x(2cos⁡x+1)cos⁡x(2cos⁡x+1)\text{RHS} = \frac{\sin x (2\cos x + 1)}{\cos x (2\cos x + 1)}

  14. Cancel the common factor (2cos⁡x+1)(2\cos x + 1), assuming 2cos⁡x+1≠02\cos x + 1 \neq 0.

    RHS=sin⁡xcos⁡x\text{RHS} = \frac{\sin x}{\cos x} RHS=tan⁡x\text{RHS} = \tan x

    This is equal to the Left Hand Side (LHS).

    LHS=RHS\text{LHS} = \text{RHS}

    Therefore, the identity is proven.

    c) Solve each of the equation for 0∘≤θ<360∘0^\circ \le \theta < 360^\circ:
    2sin⁡2θ−cos⁡θ−1=02\sin^2 \theta - \cos \theta - 1 = 0

  15. Use the identity sin⁡2θ=1−cos⁡2θ\sin^2 \theta = 1 - \cos^2 \theta to express the equation entirely in terms of cos⁡θ\cos \theta.

    2(1−cos⁡2θ)−cos⁡θ−1=02(1 - \cos^2 \theta) - \cos \theta - 1 = 0 2−2cos⁡2θ−cos⁡θ−1=02 - 2\cos^2 \theta - \cos \theta - 1 = 0 −2cos⁡2θ−cos⁡θ+1=0-2\cos^2 \theta - \cos \theta + 1 = 0

  16. Multiply by −1-1 to make the leading coefficient positive and rearrange into a standard quadratic form.

    2cos⁡2θ+cos⁡θ−1=02\cos^2 \theta + \cos \theta - 1 = 0

  17. Let y=cos⁡θy = \cos \theta and solve the quadratic equation 2y2+y−1=02y^2 + y - 1 = 0.

    Factor the quadratic:

    (2y−1)(y+1)=0(2y - 1)(y + 1) = 0

    This gives two possible values for yy:

    2y−1=0  ⟹  y=122y - 1 = 0 \implies y = \frac{1}{2} y+1=0  ⟹  y=−1y + 1 = 0 \implies y = -1

  18. Substitute back cos⁡θ\cos \theta and find the values of θ\theta in the given range 0∘≤θ<360∘0^\circ \le \theta < 360^\circ.

    Case 1: cos⁡θ=12\cos \theta = \frac{1}{2}
    The principal value is θ=60∘\theta = 60^\circ.
    Since cosine is positive in the first and fourth quadrants:

    θ=60∘\theta = 60^\circ θ=360∘−60∘=300∘\theta = 360^\circ - 60^\circ = 300^\circ

    Case 2: cos⁡θ=−1\cos \theta = -1
    The value for θ\theta is:

    θ=180∘\theta = 180^\circ

  19. List all solutions in the specified range.

    θ=60∘,180∘,300∘\boxed{\theta = 60^\circ, 180^\circ, 300^\circ}

    QUESTION FIVE

    a) Let a⃗=3i⃗−7j⃗−4k⃗\vec{a} = 3\vec{i} - 7\vec{j} - 4\vec{k}, b⃗=6i⃗+5j⃗−7k⃗\vec{b} = 6\vec{i} + 5\vec{j} - 7\vec{k} and c⃗=i⃗+j⃗−k⃗\vec{c} = \vec{i} + \vec{j} - \vec{k} be three vectors. Find:

    (i) a⃗⋅c⃗\vec{a} \cdot \vec{c}

  20. Perform the dot product of vectors a⃗\vec{a} and c⃗\vec{c}.

    a⃗⋅c⃗=(3)(1)+(−7)(1)+(−4)(−1)\vec{a} \cdot \vec{c} = (3)(1) + (-7)(1) + (-4)(-1) a⃗⋅c⃗=3−7+4\vec{a} \cdot \vec{c} = 3 - 7 + 4 a⃗⋅c⃗=0\boxed{\vec{a} \cdot \vec{c} = 0}

    (ii) b⃗×a⃗\vec{b} \times \vec{a}

  21. Perform the cross product of vectors b⃗\vec{b} and a⃗\vec{a}.

    b⃗×a⃗=∣i⃗j⃗k⃗65−73−7−4∣\vec{b} \times \vec{a} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 6 & 5 & -7 \\ 3 & -7 & -4 \end{vmatrix}

  22. Calculate the determinant.

    b⃗×a⃗=i⃗((5)(−4)−(−7)(−7))−j⃗((6)(−4)−(−7)(3))+k⃗((6)(−7)−(5)(3))\vec{b} \times \vec{a} = \vec{i}((5)(-4) - (-7)(-7)) - \vec{j}((6)(-4) - (-7)(3)) + \vec{k}((6)(-7) - (5)(3)) b⃗×a⃗=i⃗(−20−49)−j⃗(−24−(−21))+k⃗(−42−15)\vec{b} \times \vec{a} = \vec{i}(-20 - 49) - \vec{j}(-24 - (-21)) + \vec{k}(-42 - 15) b⃗×a⃗=i⃗(−69)−j⃗(−24+21)+k⃗(−57)\vec{b} \times \vec{a} = \vec{i}(-69) - \vec{j}(-24 + 21) + \vec{k}(-57) b⃗×a⃗=−69i⃗−j⃗(−3)−57k⃗\vec{b} \times \vec{a} = -69\vec{i} - \vec{j}(-3) - 57\vec{k} b⃗×a⃗=−69i⃗+3j⃗−57k⃗\boxed{\vec{b} \times \vec{a} = -69\vec{i} + 3\vec{j} - 57\vec{k}}

    (iii) ∣a⃗−b⃗−c⃗∣|\vec{a} - \vec{b} - \vec{c}|

  23. Calculate the vector a⃗−b⃗−c⃗\vec{a} - \vec{b} - \vec{c}.

    a⃗−b⃗−c⃗=(3i⃗−7j⃗−4k⃗)−(6i⃗+5j⃗−7k⃗)−(i⃗+j⃗−k⃗)\vec{a} - \vec{b} - \vec{c} = (3\vec{i} - 7\vec{j} - 4\vec{k}) - (6\vec{i} + 5\vec{j} - 7\vec{k}) - (\vec{i} + \vec{j} - \vec{k}) a⃗−b⃗−c⃗=(3−6−1)i⃗+(−7−5−1)j⃗+(−4−(−7)−(−1))k⃗\vec{a} - \vec{b} - \vec{c} = (3-6-1)\vec{i} + (-7-5-1)\vec{j} + (-4-(-7)-(-1))\vec{k} a⃗−b⃗−c⃗=(3−7)i⃗+(−13)j⃗+(−4+7+1)k⃗\vec{a} - \vec{b} - \vec{c} = (3-7)\vec{i} + (-13)\vec{j} + (-4+7+1)\vec{k} a⃗−b⃗−c⃗=−4i⃗−13j⃗+4k⃗\vec{a} - \vec{b} - \vec{c} = -4\vec{i} - 13\vec{j} + 4\vec{k}

  24. Calculate the magnitude of the resulting vector.

    ∣a⃗−b⃗−c⃗∣=(−4)2+(−13)2+(4)2|\vec{a} - \vec{b} - \vec{c}| = \sqrt{(-4)^2 + (-13)^2 + (4)^2} ∣a⃗−b⃗−c⃗∣=16+169+16|\vec{a} - \vec{b} - \vec{c}| = \sqrt{16 + 169 + 16} ∣a⃗−b⃗−c⃗∣=201\boxed{|\vec{a} - \vec{b} - \vec{c}| = \sqrt{201}}

    (iv) Find the direction cosines of the vector a⃗+b⃗\vec{a} + \vec{b}.

  25. Calculate the vector a⃗+b⃗\vec{a} + \vec{b}.

    a⃗+b⃗=(3i⃗−7j⃗−4k⃗)+(6i⃗+5j⃗−7k⃗)\vec{a} + \vec{b} = (3\vec{i} - 7\vec{j} - 4\vec{k}) + (6\vec{i} + 5\vec{j} - 7\vec{k}) a⃗+b⃗=(3+6)i⃗+(−7+5)j⃗+(−4−7)k⃗\vec{a} + \vec{b} = (3+6)\vec{i} + (-7+5)\vec{j} + (-4-7)\vec{k} a⃗+b⃗=9i⃗−2j⃗−11k⃗\vec{a} + \vec{b} = 9\vec{i} - 2\vec{j} - 11\vec{k}

  26. Calculate the magnitude of a⃗+b⃗\vec{a} + \vec{b}.

    ∣a⃗+b⃗∣=92+(−2)2+(−11)2|\vec{a} + \vec{b}| = \sqrt{9^2 + (-2)^2 + (-11)^2} ∣a⃗+b⃗∣=81+4+121|\vec{a} + \vec{b}| = \sqrt{81 + 4 + 121} ∣a⃗+b⃗∣=206|\vec{a} + \vec{b}| = \sqrt{206}

  27. Calculate the direction cosines.

    The direction cosines are x∣R⃗∣\frac{x}{|\vec{R}|}, y∣R⃗∣\frac{y}{|\vec{R}|}, z∣R⃗∣\frac{z}{|\vec{R}|}.

    cos⁡α=9206\cos \alpha = \frac{9}{\sqrt{206}} cos⁡β=−2206\cos \beta = \frac{-2}{\sqrt{206}} cos⁡γ=−11206\cos \gamma = \frac{-11}{\sqrt{206}} Direction cosines are (9206,−2206,−11206)\boxed{\text{Direction cosines are } \left(\frac{9}{\sqrt{206}}, \frac{-2}{\sqrt{206}}, \frac{-11}{\sqrt{206}}\right)}

    b) A sled is pulled by exerting a force of 100 newtons on a rope that makes a 25∘25^\circ angle with the horizontal. Find the work done in pulling the sled 40 meters.

  28. Identify the given values.

    Force F=100 NF = 100 \text{ N}
    Angle θ=25∘\theta = 25^\circ
    Distance d=40 md = 40 \text{ m}

  29. Use the formula for work done

    W=Fdcos⁡θW = F d \cos \theta.

    W=(100 N)(40 m)cos⁡25∘W = (100 \text{ N})(40 \text{ m}) \cos 25^\circ W=4000cos⁡25∘W = 4000 \cos 25^\circ

  30. Calculate the value.

    W≈4000×0.9063077W \approx 4000 \times 0.9063077 W≈3625.2308 JW \approx 3625.2308 \text{ J} W≈3625.23 J\boxed{W \approx 3625.23 \text{ J}}

    QUESTION SIX

    a) Given the polynomial g(x)=3x4−7x3+3x2−5x−1g(x) = 3x^4 - 7x^3 + 3x^2 - 5x - 1, find the remainder and quotient when g(x)g(x) is divided by (x2−2x−1)(x^2 - 2x - 1).

  31. Perform polynomial long division.

            3x^2  - x   + 2
      _________________
x^2-2x-1 | 3x^4 - 7x^3 + 3x^2 - 5x - 1
        -(3x^4 - 6x^3 - 3x^2)
        _________________
              -x^3 + 6x^2 - 5x
            -(-x^3 + 2x^2 + x)
            _________________
                    4x^2 - 6x - 1
                  -(4x^2 - 8x - 4)
                  _________________
                          2x + 3

  32. Identify the quotient and remainder.

    Quotient=3x2−x+2\boxed{\text{Quotient} = 3x^2 - x + 2} Remainder=2x+3\boxed{\text{Remainder} = 2x + 3}

    b) The polynomial f(x)=x3−5x2+ax−7f(x) = x^3 - 5x^2 + ax - 7, has the remainder −5-5 when divided by (x−1)(x-1). Find the value of aa.

  33. Apply the Remainder Theorem, which states that if a polynomial f(x)f(x) is divided by (x−c)(x-c), the remainder is f(c)f(c).

    In this case, c=1c=1 and the remainder is −5-5, so f(1)=−5f(1) = -5.

  34. Substitute x=1x=1 into the polynomial f(x)f(x).

    f(1)=(1)3−5(1)2+a(1)−7f(1) = (1)^3 - 5(1)^2 + a(1) - 7 f(1)=1−5+a−7f(1) = 1 - 5 + a - 7 f(1)=a−11f(1) = a - 11

  35. Set f(1)f(1) equal to the given remainder and solve for aa.

    a−11=−5a - 11 = -5 a=−5+11a = -5 + 11 a=6\boxed{a = 6}

    c) The motion of a particle in an electrostatic field is described by the equation, E(x)=3x3−4x2−17x+6E(x) = 3x^3 - 4x^2 - 17x + 6.

    (i) Factorise E(x)E(x).

  36. Use the Rational Root Theorem to find a rational root.

    Possible rational roots are ±pq\pm \frac{p}{q}, where pp divides 6 and qq divides 3.
    Possible roots: ±1,±2,±3,±6,±13,±23\pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{3}, \pm \frac{2}{3}.
    Test x=−2x=-2:

    E(−2)=3(−2)3−4(−2)2−17(−2)+6E(-2) = 3(-2)^3 - 4(-2)^2 - 17(-2) + 6 E(−2)=3(−8)−4(4)+34+6E(-2) = 3(-8) - 4(4) + 34 + 6 E(−2)=−24−16+34+6E(-2) = -24 - 16 + 34 + 6 E(−2)=−40+40=0E(-2) = -40 + 40 = 0

    Since E(−2)=0E(-2) = 0, (x+2)(x+2) is a factor of E(x)E(x).

  37. Perform polynomial division to find the quadratic factor.

    Divide 3x3−4x2−17x+63x^3 - 4x^2 - 17x + 6 by (x+2)(x+2).

            3x^2 - 10x + 3
      _________________
x+2 | 3x^3 - 4x^2 - 17x + 6
    -(3x^3 + 6x^2)
    _________________
          -10x^2 - 17x
        -(-10x^2 - 20x)
        _________________
                 3x + 6
               -(3x + 6)
               _________
                     0

    So, E(x)=(x+2)(3x2−10x+3)E(x) = (x+2)(3x^2 - 10x + 3).

  38. Factor the quadratic term 3x2−10x+33x^2 - 10x + 3.

    We look for two numbers that multiply to 3×3=93 \times 3 = 9 and add to −10-10. These numbers are −1-1 and −9-9.

    3x2−x−9x+33x^2 - x - 9x + 3 x(3x−1)−3(3x−1)x(3x - 1) - 3(3x - 1) (x−3)(3x−1)(x-3)(3x-1)

  39. Write the fully factorized form of E(x)E(x).

    E(x)=(x+2)(x−3)(3x−1)\boxed{E(x) = (x+2)(x-3)(3x-1)}

    (ii) Hence, solve the equation E(x)=0E(x) = 0.

  40. Set the factorized form of E(x)E(x) equal to zero.

    (x+2)(x−3)(3x−1)=0(x+2)(x-3)(3x-1) = 0

  41. Set each factor equal to zero and solve for xx.

    x+2=0  ⟹  x=−2x+2 = 0 \implies x = -2 x−3=0  ⟹  x=3x-3 = 0 \implies x = 3 3x−1=0  ⟹  3x=1  ⟹  x=133x-1 = 0 \implies 3x = 1 \implies x = \frac{1}{3} x=−2,3,13\boxed{x = -2, 3, \frac{1}{3}}
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