Past Exam Questions

Answer: \(\displaystyle \frac38p^{-2}q^{3/2}r^{-16/5}\), so \(k=\frac38\), \(a=-2\), \(b=\frac32\), \(c=-\frac{16}{5}\).

Answer: \(\displaystyle \frac38p^{-2}q^{3/2}r^{-16/5}\), so \(k=\frac38\), \(a=-2\), \(b=\frac32\), \(c=-\frac{16}{5}\).

Rewrite the expression using index notation:

Now divide powers with the same base:

\(\frac{3p}{8p^3}=\frac38p^{-2},\)

\(\frac{q^{1/2}}{q^{-1}}=q^{1/2-(-1)}=q^{3/2},\)

and

\(\frac{r^{-3}}{r^{1/5}}=r^{-3-\frac15}=r^{-16/5}.\)

Therefore the expression is

\(\frac38p^{-2}q^{3/2}r^{-16/5}.\)

Answer: \(\frac{13+5\sqrt5}{4}\).

Multiply the numerator and denominator by the conjugate of the denominator:

The denominator is

\(7^2-(3\sqrt5)^2=49-45=4.\)

The numerator is

\((4-\sqrt5)(7+3\sqrt5) =28+12\sqrt5-7\sqrt5-15.\)

So the numerator simplifies to

\(13+5\sqrt5.\)

Therefore

Answer: (a) \(\frac43\). (b) \(\frac{3\sqrt3-5}{2}\).

(a)

\(\frac{\sqrt{128}}{\sqrt{72}}=\sqrt{\frac{128}{72}}=\sqrt{\frac{16}{9}}=\frac43.\)

(b) First rationalise each term:

\(\frac{1}{1+\sqrt3}=\frac{1-\sqrt3}{1-3}=\frac{\sqrt3-1}{2}.\)

Also

\(\frac{\sqrt3}{3+2\sqrt3} =\frac{\sqrt3(3-2\sqrt3)}{(3+2\sqrt3)(3-2\sqrt3)}.\)

The denominator is \(9-12=-3\), so

\(\frac{\sqrt3}{3+2\sqrt3}=\frac{3\sqrt3-6}{-3}=2-\sqrt3.\)

Therefore

\(\frac{1}{1+\sqrt3}-\frac{\sqrt3}{3+2\sqrt3} =\frac{\sqrt3-1}{2}-(2-\sqrt3).\)

Putting this over a common denominator gives

\(\frac{\sqrt3-1-4+2\sqrt3}{2}=\frac{3\sqrt3-5}{2}.\)

Hence

\(\boxed{\frac{3\sqrt3-5}{2}}.\)

Answer: (a) \(\mathrm q(x)=3x^2-10x-8,\ r=5\); (b) \((3x+2)(x-4)(2x-5)\); (c) \(x=-\frac23,\ 4,\ \frac52\).

We start with the main method. Use polynomial division or comparison of coefficients, then apply the factor theorem to finish the factorisation.

Divide \(6x^3-35x^2+34x+45\) by \(2x-5\).

We look for

\(\mathrm p(x)=(2x-5)\mathrm q(x)+r.\)

The quotient is

\(\mathrm q(x)=3x^2-10x-8,\)

and the remainder is

\(r=5.\)

Therefore

\(\mathrm p(x)-5=(2x-5)(3x^2-10x-8).\)

Now factorise the quadratic:

\(3x^2-10x-8=(3x+2)(x-4).\)

Hence

\(\mathrm p(x)-5=(3x+2)(x-4)(2x-5).\)

For \(\mathrm p(x)=5\), set \(\mathrm p(x)-5=0\):

\((3x+2)(x-4)(2x-5)=0.\)

Thus

\(x=-\frac23,\quad x=4,\quad x=\frac52.\)

This completes the solution and gives the required result.

Answer: \(a=-2\), \(b=8\), \(\mathrm{p}(x)=(x-2)(x+1)(3x-4)\), and \(y=\frac12\ln2\) or \(y=\frac12\ln\frac43\).

Differentiate:

\(\mathrm{p}'(x)=9x^2-14x+a.\)

Using \(\mathrm{p}'(-1)=21\),

\(9+14+a=21,\)

so

\(a=-2.\)

Since \(x-2\) is a factor, \(\mathrm{p}(2)=0\):

\(24-28+2a+b=0.\)

Substitute \(a=-2\):

\(24-28-4+b=0,\)

so

\(b=8.\)

Thus

\(\mathrm{p}(x)=3x^3-7x^2-2x+8.\)

Dividing by \(x-2\) gives

\(\mathrm{p}(x)=(x-2)(3x^2-x-4).\)

Then

\(3x^2-x-4=(x+1)(3x-4),\)

so

\(\mathrm{p}(x)=(x-2)(x+1)(3x-4).\)

For the exponential equation, let \(u=\mathrm{e}^{2y}\). Then \(u\gt 0\) and

\(3u^3-7u^2-2u+8=0.\)

Using the factorisation above,

\((u-2)(u+1)(3u-4)=0.\)

Since \(u\gt 0\), we take \(u=2\) or \(u=\frac43\). Hence

\(\mathrm{e}^{2y}=2\quad\text{or}\quad \mathrm{e}^{2y}=\frac43.\)

Therefore

\(y=\frac12\ln2\quad\text{or}\quad y=\frac12\ln\frac43.\)

Answer: \(\mathrm{p}(x)=(x-1)(x-5)(x+6)\).

By the remainder theorem, the remainder when \(\mathrm{p}(x)\) is divided by \(x+2\) is \(\mathrm{p}(-2)\).

The remainder is given as \(84\), so

\(\mathrm{p}(-2)=84.\)

Now

\(\mathrm{p}(-2)=(-2)^3+A(-2)+30.\)

Thus

\(-8-2A+30=84.\)

So

\(22-2A=84,\)

which gives

\(A=-31.\)

Therefore

\(\mathrm{p}(x)=x^3-31x+30.\)

Use the factor theorem to find a linear factor. Testing \(x=1\),

\(1^3-31(1)+30=0,\)

so \((x-1)\) is a factor. Dividing gives

\(x^3-31x+30=(x-1)(x^2+x-30).\)

Finally,

\(x^2+x-30=(x-5)(x+6).\)

Hence

\(\mathrm{p}(x)=(x-1)(x-5)(x+6).\)

Answer: \(a=4\), \(\mathrm{p}(x)=2(2x+1)(4x^2+1)\), and the only real root is \(x=-\frac12\).

Use the factor theorem and the remainder theorem to turn the given information into equations for the unknown constants. After finding the constants, factorise the polynomial if required.

Since \(2x+1\) is a factor, \(x=-\frac12\) is a root. Therefore \(\mathrm{p}(-\frac12)=0\).

Substitute \(x=-\frac12\):

\(a^2\left(-\frac12\right)^3+2a\left(-\frac12\right)^2+a\left(-\frac12\right)+2=0.\)

This gives

\(-\frac{a^2}{8}+\frac{a}{2}-\frac{a}{2}+2=0,\)

so

\(-\frac{a^2}{8}+2=0.\)

Hence \(a^2=16\). Since \(a\) is positive, \(a=4\).

Now

\(\mathrm{p}(x)=16x^3+8x^2+4x+2.\)

Factorising gives

\(\mathrm{p}(x)=2(2x+1)(4x^2+1).\)

The factor \(4x^2+1\) has no real roots, since \(4x^2+1=0\) would require \(x^2=-\frac14\).

Therefore the only real root comes from \(2x+1=0\), namely \(x=-\frac12\).

The result has been obtained using the exact condition from the question, so no additional solutions are introduced.

Answer: \(a=29,\ b=-9\); \(x=-3,\frac15\); \(\theta=99.7^\circ,170.3^\circ,279.7^\circ,350.3^\circ\).

We start with the main method. Apply the factor theorem: if \((x-a)\) is a factor, then \(p(a)=0\).

(a) Since \(x+3\) is a factor of both \(\mathrm p(x)\) and \(\mathrm p'(x)\),

\(\mathrm p(-3)=0\quad\text{and}\quad \mathrm p'(-3)=0.\)

Now

\(\mathrm p'(x)=15x^2+2ax+39.\)

Using \(\mathrm p'(-3)=0\):

\(135-6a+39=0.\)

So

\(174-6a=0,\)

hence

\(a=29.\)

Using \(\mathrm p(-3)=0\):

\(5(-3)^3+29(-3)^2+39(-3)+b=0.\)

This gives

\(-135+261-117+b=0,\)

so

\(9+b=0.\)

Hence

\(b=-9.\)

(b) With these values,

\(\mathrm p(x)=5x^3+29x^2+39x-9.\)

Since \(x+3\) is a factor, divide by \(x+3\):

\(\mathrm p(x)=(x+3)(5x^2+14x-3).\)

Then

\(5x^2+14x-3=(5x-1)(x+3).\)

Thus

\(\mathrm p(x)=(x+3)^2(5x-1).\)

So the solutions of \(\mathrm p(x)=0\) are

\(x=-3\quad\text{or}\quad x=\frac15.\)

(c) Let

\(u=\operatorname{cosec}2\theta.\)

The equation becomes

\(\mathrm p(u)=0,\)

so

\(u=-3\quad\text{or}\quad u=\frac15.\)

But \(\left|\operatorname{cosec}2\theta\right|\geq1\), so \(u=\frac15\) is impossible.

Therefore

\(\operatorname{cosec}2\theta=-3,\)

which gives

\(\sin2\theta=-\frac13.\)

For \(0^\circ\leq\theta\leq360^\circ\), we have \(0^\circ\leq2\theta\leq720^\circ\).

The solutions for \(2\theta\) are approximately

\(199.47^\circ,\quad340.53^\circ,\quad559.47^\circ,\quad700.53^\circ.\)

Hence

\(\theta=99.7^\circ,\quad170.3^\circ,\quad279.7^\circ,\quad350.3^\circ.\)

This completes the solution and gives the required result.

Answer: \(p(x)=(x+4)(2x^2+3x+10)\), so the only real root is \(x=-4\).

Use the remainder theorem: when a polynomial \(P(x)\) is divided by \(x-a\), the remainder is \(P(a)\). If \(x-a\) is a factor, then \(P(a)=0\).

(a) Substitute \(x=-4\):

\(p(-4)=2(-4)^3+11(-4)^2+22(-4)+40.\)

So

\(p(-4)=-128+176-88+40=0.\)

Therefore \(x=-4\) is a root.

(b) Since \(x=-4\) is a root, \(x+4\) is a factor of \(p(x)\). Dividing \(2x^3+11x^2+22x+40\) by \(x+4\) gives

\(p(x)=(x+4)(2x^2+3x+10).\)

For the quadratic factor, the discriminant is

\(b^2-4ac=3^2-4(2)(10)=9-80=-71.\)

Since the discriminant is negative, \(2x^2+3x+10=0\) has no real roots. Hence \(p(x)=0\) has no other real roots.

Answer: (a) \(a=17\), \(b=-8\); (b) \(-8\); (c) \(34\).

Answer: (a) \(a=17\), \(b=-8\); (b) \(-8\); (c) \(34\).

Since \(p(x)\) is divisible by \(2x-1\), \(x=\frac12\) is a root. Hence

\(p\left(\frac12\right)=0.\)

So

This gives

\(\frac34+\frac{a}{4}+3+b=0.\)

Multiplying by \(4\),

\(a+4b+15=0.\)

When divided by \(x-2\), the remainder is \(p(2)\), so

\(p(2)=120.\)

Thus

\(48+4a+12+b=120,\)

which simplifies to

\(4a+b=60.\)

Solving

\(a+4b=-15,\qquad 4a+b=60\)

gives

\(a=17,\qquad b=-8.\)

The remainder when \(p(x)\) is divided by \(x\) is \(p(0)\), which is \(b\). Hence the remainder is \(-8\).

Now

\(p'(x)=18x^2+34x+6,\)

and

\(p''(x)=36x+34.\)

Therefore

\(p''(0)=34.\)

Answer: (a) \(a=19\), \(b=15\). (b) \(p(x)=(2x-3)(3x-1)(x+5)\).

Answer: (a) \(a=19\), \(b=15\). (b) \(p(x)=(2x-3)(3x-1)(x+5)\).

Differentiate \(p(x)=6x^3+ax^2-52x+b\):

\(p'(x)=18x^2+2ax-52\).

Given \(p'(1)=4\),

\(18+2a-52=4\),

so \(2a=38\) and \(a=19\).

Since \(p(x)\) is exactly divisible by \(2x-3\), \(x=\frac32\) is a root. Therefore

\(p\left(\frac32\right)=0\).

Substitute \(a=19\):

This gives

\(\frac{81}{4}+\frac{171}{4}-78+b=0\).

So

\(63-78+b=0\),

and hence \(b=15\).

Now

\(p(x)=6x^3+19x^2-52x+15\).

Dividing by \(2x-3\) gives

\(p(x)=(2x-3)(3x^2+14x-5)\).

Finally,

\(3x^2+14x-5=(3x-1)(x+5)\).

Therefore

\(p(x)=(2x-3)(3x-1)(x+5)\).

Answer: (b) \(a=6,\ c=-10\); (c) \(q(x)=2x^2+3x-5\); (d) \(p(x)=(3x+2)(x-1)(2x+5)\).

Answer: (b) \(a=6,\ c=-10\); (c) \(q(x)=2x^2+3x-5\); (d) \(p(x)=(3x+2)(x-1)(2x+5)\).

(a) Differentiate

\(p(x)=ax^3+13x^2+bx+c.\)

This gives

\(p'(x)=3ax^2+26x+b.\)

Therefore

\(p'(0)=b.\)

Since \(p'(0)=-9\),

\(b=-9.\)

(b) Since \(3x+2\) is a factor,

\(p\left(-\frac23\right)=0.\)

Using \(b=-9\),

\(p(x)=ax^3+13x^2-9x+c.\)

Substituting \(x=-\frac23\):

\(-\frac{8a}{27}+\frac{52}{9}+6+c=0.\)

Multiplying by \(27\),

\(-8a+318+27c=0.\)

So

\(8a-27c=318.\)

Also, when \(p(x)\) is divided by \(x+1\), the remainder is \(6\). Hence

\(p(-1)=6.\)

This gives

\(-a+13+9+c=6.\)

So

\(a-c=16.\)

Solving

\(8a-27c=318,\qquad a-c=16,\)

gives

\(a=6,\qquad c=-10.\)

(c) Therefore

\(p(x)=6x^3+13x^2-9x-10.\)

Dividing by \(3x+2\) gives

\(q(x)=2x^2+3x-5.\)

(d) Factorise the quadratic:

\(2x^2+3x-5=(x-1)(2x+5).\)

Hence

\(p(x)=(3x+2)(x-1)(2x+5).\)

Answer: \(a=-27\), \(b=3\); \(\mathrm p(x)=(2x+1)(5x-1)(x-3)\); remainder \(3\).

Since \(\mathrm p(x)\) is divisible by \(2x+1\), the factor theorem gives \(\mathrm p\left(-\frac12\right)=0\).

So

\(-\frac54+\frac a4+5+b=0.\)

Multiplying by \(4\),

\(a+4b=-15. \tag{1}\)

The remainder when \(\mathrm p(x)\) is divided by \(x+1\) is \(\mathrm p(-1)\). Hence

\(\mathrm p(-1)=-24.\)

So

\(10(-1)^3+a(-1)^2-10(-1)+b=-24,\)

which gives

\(a+b=-24. \tag{2}\)

Subtracting (2) from (1),

\(3b=9,\)

so \(b=3\). Then \(a=-27\).

Thus

\(\mathrm p(x)=10x^3-27x^2-10x+3.\)

Since \(2x+1\) is a factor, division gives

\(\mathrm p(x)=(2x+1)(5x^2-16x+3).\)

Factorising the quadratic,

\(5x^2-16x+3=(5x-1)(x-3).\)

Therefore

\(\mathrm p(x)=(2x+1)(5x-1)(x-3).\)

The remainder when \(\mathrm p(x)\) is divided by \(x\) is \(\mathrm p(0)\), so the remainder is \(3\).

Answer: normal \(y-3=\frac1{11}(x-1)\); \(a=2\); \(x=-2\) or \(x=\frac32\).

(a) Differentiate

\(\mathrm f(x)=4x^3-4x^2-15x+18.\)

This gives

\(\mathrm f'(x)=12x^2-8x-15.\)

When \(x=1\),

\(\mathrm f(1)=4-4-15+18=3\)

and

\(\mathrm f'(1)=12-8-15=-11.\)

So the tangent gradient is \(-11\), and the normal gradient is

\(\frac1{11}.\)

The normal passes through \((1,3)\), so its equation is

\(y-3=\frac1{11}(x-1).\)

(b) Since \(x+a\) is a factor, a root is \(x=-a\). Test \(x=-2\):

\(\mathrm f(-2)=4(-8)-4(4)-15(-2)+18.\)

So

\(\mathrm f(-2)=-32-16+30+18=0.\)

Therefore \(x+2\) is a factor, so

\(a=2.\)

Divide \(4x^3-4x^2-15x+18\) by \(x+2\):

\(\mathrm f(x)=(x+2)(4x^2-12x+9).\)

The quadratic factor is a square:

\(4x^2-12x+9=(2x-3)^2.\)

Thus

\(\mathrm f(x)=(x+2)(2x-3)^2.\)

Hence

\(x=-2\quad\text{or}\quad 2x-3=0.\)

Therefore

\(x=-2\quad\text{or}\quad x=\frac32.\)

Answer: \(a=6\), \(b=-24\); \(p(x)=(2x+1)(3x^2-12)\); \(x=-\dfrac12,\pm2\).

(a) Since \(2x+1\) is a factor, \(x=-\frac12\) is a root.

So

\(p\left(-\frac12\right)=0.\)

This gives

\(a\left(-\frac18\right)+3\left(\frac14\right)+b\left(-\frac12\right)-12=0.\)

Multiplying by 8,

\(-a+6-4b-96=0.\)

Hence

\(a+4b=-90.\)

The remainder when \(p(x)\) is divided by \(x-3\) is \(p(3)\), so

\(p(3)=105.\)

Thus

\(27a+27+3b-12=105.\)

So

\(9a+b=30.\)

Solving

\(a+4b=-90,\qquad 9a+b=30\)

gives

\(a=6,\qquad b=-24.\)

(b) Then

\(p(x)=6x^3+3x^2-24x-12.\)

Dividing by \(2x+1\),

\(p(x)=(2x+1)(3x^2-12).\)

(c) Solve

\((2x+1)(3x^2-12)=0.\)

So

\(2x+1=0 \quad\text{or}\quad 3x^2-12=0.\)

Therefore

\(x=-\frac12,\qquad x=\pm2.\)

Answer: \(Q(x)=6x^2-11x-10\); \(x=2,-\dfrac23,\dfrac52\).

By the remainder theorem, the remainder when \(p(x)\) is divided by \(x-3\) is \(p(3)\). So

\(p(3)=11.\)

Substitute \(x=3\):

\(162+9a+36+b=11.\)

Hence

\(9a+b=-187.\)

The remainder when \(p(x)\) is divided by \(x+1\) is \(p(-1)\). So

\(p(-1)=-21.\)

Substitute \(x=-1\):

\(-6+a-12+b=-21.\)

Hence

\(a+b=-3.\)

Solving the two equations

\(9a+b=-187,\qquad a+b=-3\)

gives

\(a=-23,\qquad b=20.\)

Therefore

\(p(x)=6x^3-23x^2+12x+20.\)

Since \(p(x)=(x-2)Q(x)\), divide \(6x^3-23x^2+12x+20\) by \(x-2\):

\(Q(x)=6x^2-11x-10.\)

(b) Factorise \(Q(x)\):

\(6x^2-11x-10=(3x+2)(2x-5).\)

So

\(p(x)=(x-2)(3x+2)(2x-5).\)

Hence \(p(x)=0\) gives

\(x=2,\qquad x=-\frac23,\qquad x=\frac52.\)

Answer: remainder \(24\); \(p(x)=(x+2)(3x-1)(5x-1)\).

(a) By the remainder theorem, the remainder when \(p(x)\) is divided by \(x+1\) is \(p(-1)\).

\(p(-1)=15(-1)^3+22(-1)^2-15(-1)+2.\)

So

\(p(-1)=-15+22+15+2=24.\)

The remainder is therefore \(24\).

(b)(i) To show that \(x+2\) is a factor, evaluate \(p(-2)\):

\(p(-2)=15(-8)+22(4)-15(-2)+2.\)

Thus

\(p(-2)=-120+88+30+2=0.\)

Since \(p(-2)=0\), \(x+2\) is a factor of \(p(x)\).

(b)(ii) Divide \(p(x)\) by \(x+2\):

\(p(x)=(x+2)(15x^2-8x+1).\)

Now factorise the quadratic:

\(15x^2-8x+1=(3x-1)(5x-1).\)

Therefore

\(p(x)=(x+2)(3x-1)(5x-1).\)

Answer: \(\mathrm{p}(x)=(x+4)(6x^2-5x+1)=(x+4)(3x-1)(2x-1)\); remainder \(-19\).

Since \(x+4\) is a factor,

\(\mathrm{p}(-4)=0.\)

Therefore

\(-64a+16b+76+4=0,\)

so

\(-4a+b+5=0.\)

Also

\(2\mathrm{p}(1)=5\mathrm{p}(0).\)

Now

\(\mathrm{p}(1)=a+b-19+4=a+b-15\)

and

\(\mathrm{p}(0)=4.\)

Thus

\(2(a+b-15)=20,\)

which gives

\(a+b=25.\)

Solving

\(-4a+b+5=0,\qquad a+b=25,\)

gives

\(a=6,\qquad b=19.\)

So

\(\mathrm{p}(x)=6x^3+19x^2-19x+4.\)

Dividing by \(x+4\),

\(\mathrm{p}(x)=(x+4)(6x^2-5x+1).\)

Hence

\(A=6,\qquad B=-5,\qquad C=1.\)

The quadratic factorises as

\(6x^2-5x+1=(3x-1)(2x-1).\)

Therefore

\(\mathrm{p}(x)=(x+4)(3x-1)(2x-1).\)

Now

\(\mathrm{p}'(x)=18x^2+38x-19.\)

When a polynomial is divided by \(x\), the remainder is its constant term, so the remainder is

\(-19.\)

Answer: \(\mathrm{p}\left(\frac12\right)=0\); \(\mathrm{p}(x)=(2x-1)(x-4)(x+3)\); \(x=\frac12,4,-3\).

(a) Substitute \(x=\frac12\):

\(\mathrm{p}\left(\frac12\right) =2\left(\frac18\right)-3\left(\frac14\right)-23\left(\frac12\right)+12.\)

So

\(\mathrm{p}\left(\frac12\right) =\frac14-\frac34-\frac{23}{2}+12=0.\)

(b) Since \(\mathrm{p}\left(\frac12\right)=0\), \(2x-1\) is a factor.

Dividing \(2x^3-3x^2-23x+12\) by \(2x-1\) gives

\(x^2-x-12.\)

Therefore

\(\mathrm{p}(x)=(2x-1)(x^2-x-12).\)

Factorise the quadratic:

\(x^2-x-12=(x-4)(x+3).\)

Hence

\(\mathrm{p}(x)=(2x-1)(x-4)(x+3).\)

Solving \(\mathrm{p}(x)=0\),

\(x=\frac12,\qquad x=4,\qquad x=-3.\)

Answer: \(x=\frac12\), \(x=1\), \(x=3\).

At the points where the curve cuts the \(x\)-axis, \(y=0\). Therefore

\((2x-9)(x^2+5)+42=0.\)

Expand the expression:

\((2x-9)(x^2+5)+42=2x^3-9x^2+10x-45+42.\)

So

\(2x^3-9x^2+10x-3=0.\)

Use the factor theorem. Testing \(x=1\),

\(2(1)^3-9(1)^2+10(1)-3=0,\)

so \((x-1)\) is a factor. Dividing the cubic by \((x-1)\) gives

\(2x^3-9x^2+10x-3=(x-1)(2x^2-7x+3).\)

Now factorise the quadratic:

\(2x^2-7x+3=(2x-1)(x-3).\)

Hence

\((x-1)(2x-1)(x-3)=0.\)

Therefore the \(x\)-coordinates are

\(x=\frac12,\quad x=1,\quad x=3.\)