Past Exam Questions

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: \(a=27\), \(b=84\); \(p(x)=(2x-1)(x^2+14x+49)=(2x-1)(x+7)^2\).

Differentiate \(p(x)\):

\(p'(x)=6x^2+2ax+b.\)

Since the remainder when \(p'(x)\) is divided by \(x+3\) is \(-24\),

\(p'(-3)=-24.\)

So

\(54-6a+b=-24,\)

which gives

\(6a-b=78.\)

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

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

This gives

\(\frac14+\frac a4+\frac b2-49=0.\)

Multiplying by 4,

\(a+2b=195.\)

Solving

\(6a-b=78,\qquad a+2b=195\)

gives

\(a=27,\qquad b=84.\)

So

\(p(x)=2x^3+27x^2+84x-49.\)

Dividing by \(2x-1\),

\(p(x)=(2x-1)(x^2+14x+49).\)

Finally, \(x^2+14x+49=(x+7)^2\), so

\(\boxed{p(x)=(2x-1)(x+7)^2}.\)

Answer: \(a=4\); \((x-2)(2x-3)(2x+1)\); \(x=2,\dfrac32,-\dfrac12\).

Since \(x-2\) is a factor, substituting \(x=2\) must give zero.

\(a(2)^3-12(2)^2+5(2)+6=0.\)

Thus

\(8a-48+10+6=0,\)

so \(8a-32=0\), giving

\(\boxed{a=4}.\)

Now factorise \(4x^3-12x^2+5x+6\). Since \(x-2\) is a factor, divide by \(x-2\):

\(4x^3-12x^2+5x+6=(x-2)(4x^2-4x-3).\)

Then

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

Therefore

\(4x^3-12x^2+5x+6=(x-2)(2x-3)(2x+1).\)

Hence the solutions are

\(\boxed{x=2,\quad x=\frac32,\quad x=-\frac12}.\)

Answer: \(a=-23\), \(b=-30\); \(p(x)=(x+4)(2x+1)(x-3)\), so \(x=-4,-\frac12,3\).

(i) Since \(x+4\) is a factor, \(p(-4)=0\).

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

Thus

\(-128+48-4a-12=0.\)

So

\(-92-4a=0, \qquad a=-23.\)

The remainder when \(p(x)\) is divided by \(x-1\) is \(p(1)\). With \(a=-23\),

\(p(1)=2+3-23-12=-30.\)

Hence \(b=-30\).

(ii) Now

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

Since \(x+4\) is a factor, divide by \(x+4\):

\(2x^3+3x^2-23x-12=(x+4)(2x^2-5x-3).\)

Factorise the quadratic:

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

Therefore

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

So \(p(x)=0\) when

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

Answer: \(a=-23\), \(b=-30\); \(p(x)=(x+4)(2x+1)(x-3)\), so \(x=-4,-\frac12,3\).

(i) Since \(x+4\) is a factor, \(p(-4)=0\).

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

Thus

\(-128+48-4a-12=0.\)

So

\(-92-4a=0, \qquad a=-23.\)

The remainder when \(p(x)\) is divided by \(x-1\) is \(p(1)\). With \(a=-23\),

\(p(1)=2+3-23-12=-30.\)

Hence \(b=-30\).

(ii) Now

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

Since \(x+4\) is a factor, divide by \(x+4\):

\(2x^3+3x^2-23x-12=(x+4)(2x^2-5x-3).\)

Factorise the quadratic:

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

Therefore

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

So \(p(x)=0\) when

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

Answer: (i) \(a=b=6\). (ii) \(p(x)=(2x-1)(3x^2+10x+8)\). (iii) \(p(x)=(2x-1)(x+2)(3x+4)\). (iv) \(\theta=30^\circ,\ 150^\circ\).

Since \(p(x)\) is divisible by \(2x-1\), \(x=\dfrac12\) is a root. Therefore

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

This gives

\(a\left(\frac12\right)^3 +17\left(\frac12\right)^2 +b\left(\frac12\right)-8=0.\)

So

\(\frac{a}{8}+\frac{17}{4}+\frac{b}{2}-8=0.\)

The remainder when \(p(x)\) is divided by \(x+3\) is \(p(-3)\). Hence

\(p(-3)=-35.\)

This gives

\(-27a+153-3b-8=-35.\)

From (1), multiply by \(8\):

\(a+4b=30.\)

From (2),

\(-27a-3b+145=-35,\)

so

\(9a+b=60.\)

Solving \(a+4b=30\) and \(9a+b=60\) gives

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

Thus \(a=b\), as required.

Now

\(p(x)=6x^3+17x^2+6x-8.\)

Dividing by \(2x-1\),

\(p(x)=(2x-1)(3x^2+10x+8).\)

Factorise the quadratic:

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

So

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

For the trigonometric equation, substitute \(a=b=6\):

\(6\sin^3\theta+17\sin^2\theta+6\sin\theta-8=0.\)

This is \(p(\sin\theta)=0\), so

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

Only \(\sin\theta=\dfrac12\) is possible, since \(\sin\theta=-2\) and \(\sin\theta=-\dfrac43\) are outside the range of sine.

For \(0^\circ\lt \theta\lt 180^\circ\),

\(\sin\theta=\frac12\)

gives

\(\theta=30^\circ,\ 150^\circ.\)

Answer: \(a=-27\), \(b=9\); \(p(x)=(2x+1)(x^2-14x+9)\); \(x=-\frac12,\ 7\pm2\sqrt{10}\).

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

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

Substitute:

\(2\left(-\frac12\right)^3+a\left(-\frac12\right)^2+4\left(-\frac12\right)+b=0.\)

This gives

\(-\frac14+\frac a4-2+b=0,\)

so

\(a+4b=9.\)

Also, when \(p(x)\) is divided by \(x-1\), the remainder is \(-12\). By the remainder theorem,

\(p(1)=-12.\)

So

\(2+a+4+b=-12,\)

hence

\(a+b=-18.\)

Solve the simultaneous equations:

\(a+4b=9,\qquad a+b=-18.\)

Subtract the second from the first:

\(3b=27\Rightarrow b=9.\)

Then

\(a=-18-9=-27.\)

(ii) So

\(p(x)=2x^3-27x^2+4x+9.\)

Divide by \(2x+1\) or factor by inspection:

\(p(x)=(2x+1)(x^2-14x+9).\)

(iii) Therefore

\(p(x)=0 \quad\Longrightarrow\quad (2x+1)(x^2-14x+9)=0.\)

So either

\(2x+1=0\Rightarrow x=-\frac12,\)

or

\(x^2-14x+9=0.\)

Using the quadratic formula,

\(x=\frac{14\pm\sqrt{14^2-4\cdot1\cdot9}}{2} = \frac{14\pm\sqrt{160}}{2} = \frac{14\pm4\sqrt{10}}{2} = 7\pm2\sqrt{10}.\)

Hence the exact solutions are

\(x=-\frac12,\qquad x=7\pm2\sqrt{10}.\)

Answer: \((x-2)(2x-1)(5x+2)\).

We test a simple integer root. Substituting \(x=2\):

\(10(2)^3-21(2)^2+4=80-84+4=0.\)

So \(x-2\) is a factor.

Dividing \(10x^3-21x^2+0x+4\) by \(x-2\) gives

\(10x^2-x-2.\)

Now factorise the quadratic:

\(10x^2-x-2 = 10x^2-5x+4x-2 = 5x(2x-1)+2(2x-1) = (5x+2)(2x-1).\)

Hence the cubic factorises as

\(\boxed{(x-2)(2x-1)(5x+2)}.\)

Answer: \(a=12\), \(b=4\); remainder \(9\).

Since \(2x-1\) is a factor, \(x=\dfrac12\) is a root of both \(p(x)\) and \(p'(x)\).

First use \(p\left(\dfrac12\right)=0\):

\(\dfrac{a}{8}+\dfrac{b}{4}-\dfrac{13}{2}+4=0.\)

Multiply by \(8\):

\(a+2b-52+32=0,\)

so

\(a+2b=20.\)

Now

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

Use \(p'\left(\dfrac12\right)=0\):

\(\dfrac{3a}{4}+b-13=0.\)

Multiply by \(4\):

\(3a+4b=52.\)

Solving

\(a+2b=20,\qquad 3a+4b=52,\)

gives

\(\boxed{a=12,\quad b=4}.\)

So

\(p(x)=12x^3+4x^2-13x+4.\)

The remainder on division by \(x+1\) is \(p(-1)\):

\(p(-1)=-12+4+13+4=9.\)

Therefore the remainder is \(\boxed{9}\).

Answer: \(a=12\), \(b=-17\); \(x=0,\ -\dfrac13\pm\dfrac{\sqrt{55}}6\).

Since \(2x-1\) is a factor, \(x=\dfrac12\) is a root. So

\(p\left(\dfrac12\right)=0.\)

This gives

\(\dfrac{a}{8}+8\left(\dfrac14\right)+\dfrac b2+5=0.\)

So

\(\dfrac a8+\dfrac b2+7=0.\)

Multiply by \(8\):

\(a+4b+56=0.\)

The remainder when divided by \(x+2\) is \(p(-2)\). Therefore

\(-8a+8(4)-2b+5=-25.\)

So

\(-8a-2b+37=-25,\)

which gives

\(4a+b=31.\)

Solving

\(a+4b=-56,\qquad 4a+b=31,\)

gives

\(a=12,\qquad b=-17.\)

Now

\(p(x)=12x^3+8x^2-17x+5.\)

For \(p(x)=5\),

\(12x^3+8x^2-17x=0.\)

Factor out \(x\):

\(x(12x^2+8x-17)=0.\)

So \(x=0\), or

\(12x^2+8x-17=0.\)

Using the quadratic formula,

\(x=\dfrac{-8\pm\sqrt{8^2-4(12)(-17)}}{24}=\dfrac{-8\pm\sqrt{880}}{24}.\)

Since \(\sqrt{880}=4\sqrt{55}\),

\(x=-\dfrac13\pm\dfrac{\sqrt{55}}6.\)

Therefore

\(\boxed{x=0,\ -\dfrac13+\dfrac{\sqrt{55}}6,\ -\dfrac13-\dfrac{\sqrt{55}}6}.\)

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

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