Past Exam Questions

Answer: \(a=2\), \(b=2\); \(x=-1,\frac12,2\).

Work through the given conditions step by step, keeping exact values until the final answer is required.

Since \(x=2\) is a root,

\(8a-12-6+b=0\), so \(8a+b=18\).

Since \(x=-1\) is a root,

\(-a-3+3+b=0\), so \(b=a\).

Therefore \(9a=18\), giving \(a=2\), and hence \(b=2\).

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

The known roots give factors \((x-2)\) and \((x+1)\). Dividing or factorising gives

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

Thus the solutions are \(x=2\), \(x=-1\), and \(x=\frac12\).

This gives the required answer: \(a=2\), \(b=2\); \(x=-1,\frac12,2\).

Answer: (a) \(33\); (b)(ii) \((2x-1)(3x+5)(x-1)\); (b)(iii) \(\theta=30^\circ,90^\circ\).

We start with the main method. Use the factor theorem first, then factorise the remaining quadratic factor.

For part (a), by the remainder theorem the remainder on division by \(x-2\) is \(\mathrm p(2)\).

\(\mathrm p(2)=6(2)^3+(2)^2-12(2)+5=48+4-24+5=33.\)

For part (b)(i), evaluate \(\mathrm p\left(\frac12\right)\):

\(\mathrm p\left(\frac12\right)=6\left(\frac12\right)^3+\left(\frac12\right)^2-12\left(\frac12\right)+5.\)

This is

\(\frac34+\frac14-6+5=0.\)

Therefore \(2x-1\) is a factor.

Dividing \(6x^3+x^2-12x+5\) by \(2x-1\) gives

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

Now factorise:

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

Hence

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

For part (b)(iii), let \(u=\sin\theta\). Then

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

has factors

\((2u-1)(3u+5)(u-1)=0.\)

So

\(u=\frac12,\quad u=-\frac53,\quad u=1.\)

Since \(0^\circ\leqslant\theta\leqslant90^\circ\), \(u=-\frac53\) is impossible. Therefore

\(\sin\theta=\frac12\) gives \(\theta=30^\circ\), and \(\sin\theta=1\) gives \(\theta=90^\circ\).

This completes the solution and gives the required result.

Answer: \(x=-3,\frac12,4\), giving \(A(-3,-23)\), \(B\left(\frac12,12\right)\), \(C(4,47)\). The midpoint of \(AC\) is \(B\).

Work through the problem step by step, using exact values where possible before giving any final numerical approximation.

(a) To show that \(x+3\) is a factor, substitute \(x=-3\) into the expression:

\(-12+23(-3)+3(-3)^2-2(-3)^3.\)

This is

\(-12-69+27+54=0.\)

Therefore \(x+3\) is a factor.

(b) At the intersections,

\(-5+33x+3x^2-2x^3=10x+7.\)

Rearranging gives

\(-12+23x+3x^2-2x^3=0.\)

Using part (a), factorise this cubic:

\(-12+23x+3x^2-2x^3=-(x+3)(2x-1)(x-4).\)

Hence the \(x\)-coordinates of the three intersection points are

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

The corresponding \(y\)-values can be found from the line \(y=10x+7\):

\(x=-3:\quad y=-30+7=-23,\)

\(x=\frac12:\quad y=5+7=12,\)

\(x=4:\quad y=40+7=47.\)

So

\(A(-3,-23),\quad B\left(\frac12,12\right),\quad C(4,47).\)

The midpoint of \(AC\) is

\(\left(\frac{-3+4}{2},\frac{-23+47}{2}\right) =\left(\frac12,12\right).\)

This is exactly the point \(B\), so \(B\) is the midpoint of \(AC\).

Answer: (a) \(f(x)=(x-1)(x-4)\) or \(f(x)=-(x-1)(x-4)\); (b) \(p(x)=5x^3+14x^2+7x-2\).

Answer: (a) \(f(x)=(x-1)(x-4)\) or \(f(x)=-(x-1)(x-4)\); (b) \(p(x)=5x^3+14x^2+7x-2\).

(a) From the graph of \(y=|f(x)|\), the zeros occur at \(x=1\) and \(x=4\). Hence \(f(x)\) must have the factors \((x-1)\) and \((x-4)\).

Also, when \(x=0\), the graph has value \(4\), so

\(|f(0)|=4.\)

If \(f(x)=k(x-1)(x-4)\), then

\(|f(0)|=|4k|=4,\)

so \(k=1\) or \(k=-1\). Therefore

\(f(x)=(x-1)(x-4) \quad\text{or}\quad f(x)=-(x-1)(x-4).\)

(b) Since the roots are \(\frac15\), \(n\) and \(n+1\), and the leading coefficient is \(5\),

\(p(x)=(5x-1)(x-n)(x-(n+1)).\)

For \(5x^3+ax^2+bx-2\), the product of the roots is

\(-\frac{-2}{5}=\frac25.\)

Thus

\(\frac15 n(n+1)=\frac25.\)

So

\(n(n+1)=2.\)

This gives

\(n^2+n-2=0,\)

or

\((n+2)(n-1)=0.\)

Since \(n\) is negative, \(n=-2\). Hence the roots are \(\frac15\), \(-2\) and \(-1\), so

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

Now

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

Therefore

\(p(x)=(5x-1)(x^2+3x+2)=5x^3+14x^2+7x-2.\)

Answer: (b) \((x-1)(x+4)(x-5)\); (c) \(y=0\) or \(y=\ln5\).

Answer: (b) \((x-1)(x+4)(x-5)\); (c) \(y=0\) or \(y=\ln5\).

(a) Substitute \(x=1\) into the polynomial:

\(1^3-2(1)^2-19(1)+20=1-2-19+20=0.\)

Therefore \(x-1\) is a factor.

(b) Divide \(x^3-2x^2-19x+20\) by \(x-1\), or compare coefficients:

\(x^3-2x^2-19x+20=(x-1)(x^2-x-20).\)

Then factorise the quadratic:

\(x^2-x-20=(x+4)(x-5).\)

So

\(x^3-2x^2-19x+20=(x-1)(x+4)(x-5).\)

(c) Let \(X=\mathrm e^y\). Then \(X\gt 0\), and the equation becomes

\(X^3-2X^2-19X+20=0.\)

Using the factorisation from part (b),

\((X-1)(X+4)(X-5)=0.\)

Thus

\(X=1,\quad X=-4,\quad X=5.\)

Since \(X=\mathrm e^y\gt 0\), reject \(X=-4\). Hence

\(\mathrm e^y=1\quad\text{or}\quad \mathrm e^y=5.\)

Therefore

\(y=0\quad\text{or}\quad y=\ln5.\)

Answer: \(\mathrm p(x)=(4x+1)(3x-1)^2\), so \(x=\frac13\) is a repeated root.

(a) Substitute \(x=-0.25=-\frac14\):

This is

\(-\frac{36}{64}-\frac{15}{16}+\frac12+1=0.\)

So \(x=-0.25\) is a root.

(b) Since \(x=-\frac14\) is a root, \(4x+1\) is a factor.

Factorising the polynomial gives

\(36x^3-15x^2-2x+1=(4x+1)(9x^2-6x+1).\)

The quadratic factor is

\(9x^2-6x+1=(3x-1)^2.\)

Therefore

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

So \(x=\frac13\) is a repeated root of \(\mathrm p(x)=0\).

Answer: \(p(x)=2x^3-x^2-8x+4\).

Since the roots are \(\frac12\), \(n\) and \(-n\), write

\(p(x)=2\left(x-\frac12\right)(x-n)(x+n).\)

This is

\(p(x)=(2x-1)(x^2-n^2).\)

Expanding gives

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

The \(y\)-intercept is \(p(0)\), so

\(p(0)=n^2=4.\)

Hence \(n^2=4\). Therefore

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

So

\(\boxed{p(x)=2x^3-x^2-8x+4}.\)

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

The graph crosses the \(x\)-axis at

\(x=-2,\quad x=-1,\quad x=4.\)

So the cubic must have the form

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

From the diagram, the curve crosses the \(y\)-axis at \(24\) or \(-24\), depending on which of the two possible orientations is used. When \(x=0\),

\((x+2)(x+1)(x-4)=2\cdot1\cdot(-4)=-8.\)

Hence \(-8k=24\) gives \(k=-3\), and \(-8k=-24\) gives \(k=3\).

Therefore the two possible expressions are

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

or

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

Answer: \(\mathrm{f}(x)=-\frac12(x+5)(x+1)(x-2)\); \(-5\lt x\lt -1\) or \(x\gt 2\).

(a) From the graph, the roots are

\(x=-5,\quad x=-1,\quad x=2.\)

Therefore

\(\mathrm{f}(x)=k(x+5)(x+1)(x-2).\)

The graph cuts the \(y\)-axis at \(5\), so \(\mathrm{f}(0)=5\). Hence

\(k(5)(1)(-2)=5.\)

So

\(-10k=5,\qquad k=-\frac12.\)

Thus

\(\mathrm{f}(x)=-\frac12(x+5)(x+1)(x-2).\)

(b) The graph is below the \(x\)-axis where \(\mathrm{f}(x)\lt 0\). From the sketch this happens between the first two roots and to the right of the last root:

\(-5\lt x\lt -1\quad\text{or}\quad x\gt 2.\)

Answer: (a) other root \(5\), \(a=-11\), \(b=38\). (b) \(x=-4,-1,10\).

(a) For a monic cubic with roots \(2\), \(4\), and \(p\), the product of the roots is \(40\), since the constant term is \(-40\).

Thus

\(2\cdot4\cdot p=40,\)

so

\(p=5.\)

The sum of the roots is

\(2+4+5=11.\)

Since the coefficient of \(x^2\) is the negative of the sum of the roots,

\(a=-11.\)

The coefficient of \(x\) is the sum of the pairwise products:

\(b=2\cdot4+2\cdot5+4\cdot5=8+10+20=38.\)

(b) Since there is one positive integer root and two negative integer roots, try factors of \(40\). The factorisation is

\(x^3-5x^2-46x-40=(x-10)(x+4)(x+1).\)

Therefore the roots are

\(\boxed{x=10,\ -4,\ -1}.\)

Answer: \(p=-4\), \(a=-1\), \(b=-14\), \(\frac{dy}{dx}=3x^2-2x-14\), \(x=-2\), \(y=40\), tangent \(y=2x+44\).

For a monic cubic, the product of the roots is the negative of the constant term. Hence

\(2\cdot3\cdot p=-24,\)

so

\(p=-4.\)

The polynomial is therefore

\((x-2)(x-3)(x+4).\)

First,

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

Then

\((x^2-5x+6)(x+4)=x^3-x^2-14x+24.\)

So

\(a=-1,\qquad b=-14.\)

For the curve \(y=x^3-x^2-14x+24\),

\(\frac{dy}{dx}=3x^2-2x-14.\)

Set the gradient equal to \(2\):

\(3x^2-2x-14=2.\)

So

\(3x^2-2x-16=0.\)

Solving gives

\(x=\frac{2\pm14}{6}.\)

The integer value is

\(x=-2.\)

The corresponding value of \(y\) is

\(y=(-2)^3-(-2)^2-14(-2)+24=40.\)

At \(P(-2,40)\), the tangent gradient is \(2\). Hence

\(y-40=2(x+2),\)

so

\(\boxed{y=2x+44}.\)

Answer: (i) other root \(4\), \(a=-10\), \(b=33\). (ii) \(6,6,1\); \(2,2,9\); \(1,1,36\).

For a monic cubic \(x^3+ax^2+bx-36=0\), the product of the three roots is \(36\).

If \(x=3\) is the repeated root, the roots are \(3\), \(3\), and another root \(r\).

So

\(3\cdot3\cdot r=36.\)

Hence

\(r=4.\)

The cubic is therefore

\((x-3)^2(x-4).\)

Expand:

\((x^2-6x+9)(x-4)=x^3-10x^2+33x-36.\)

Thus

\(a=-10,\qquad b=33.\)

For the other possibilities, let the repeated positive integer root be \(r\) and the third integer root be \(s\).

Then

\(r^2s=36.\)

The possible repeated positive integer roots, excluding \(r=3\), are \(r=1\), \(r=2\), and \(r=6\).

If \(r=1\), then \(s=36\), so the roots are \(1,1,36\).

If \(r=2\), then \(s=9\), so the roots are \(2,2,9\).

If \(r=6\), then \(s=1\), so the roots are \(6,6,1\).

Therefore the other possible sets of roots are

\(\boxed{6,6,1;\quad 2,2,9;\quad 1,1,36}.\)

Answer: \(a=2\), \(b=0\).

Since \(x-3\) is a factor of \(\mathrm{p}(x)\), the factor theorem gives

\(\mathrm{p}(3)=0\).

Substitute \(x=3\) into

\(\mathrm{p}(x)=ax^3-7x^2-bx+9\):

\(27a-7(9)-3b+9=0\).

This simplifies to

\(27a-3b-54=0\).

Dividing by \(3\),

\(9a-b=18\). \(\quad\text{(1)}\)

When \(\mathrm{p}(x)\) is divided by \(x+2\), the remainder is \(\mathrm{p}(-2)\). The remainder is given as \(-35\), so

\(\mathrm{p}(-2)=-35\).

Substituting \(x=-2\),

\(-8a-7(4)+2b+9=-35\).

So

\(-8a+2b-19=-35\),

which gives

\(-8a+2b=-16\).

Dividing by \(2\),

\(-4a+b=-8\). \(\quad\text{(2)}\)

Add equations (1) and (2):

\((9a-b)+(-4a+b)=18-8\).

Thus

\(5a=10\),

so

\(a=2\).

Substitute into \(-4a+b=-8\):

\(-4(2)+b=-8\),

so

\(b=0\).

Answer: \(a=2\), \(b=-1\).

Since \(x+2\) is a factor of \(\mathrm p(x)\), the factor theorem gives

\(\mathrm p(-2)=0.\)

Substitute \(x=-2\) into \(\mathrm p(x)=x^3+ax^2+bx-2\):

\((-2)^3+a(-2)^2+b(-2)-2=0.\)

So

\(-8+4a-2b-2=0,\)

which simplifies to

\(4a-2b=10.\)

Divide by \(2\):

\(2a-b=5. \tag{1}\)

By the remainder theorem, the remainder when \(\mathrm p(x)\) is divided by \(x-3\) is \(\mathrm p(3)\). This remainder is \(40\), so

\(3^3+a(3)^2+b(3)-2=40.\)

Therefore

\(27+9a+3b-2=40,\)

so

\(9a+3b=15.\)

Divide by \(3\):

\(3a+b=5. \tag{2}\)

From (1),

\(b=2a-5.\)

Substitute this into (2):

\(3a+(2a-5)=5.\)

So

\(5a=10,\)

giving

\(a=2.\)

Then

\(b=2(2)-5=-1.\)

Therefore

\(a=2,\qquad b=-1.\)

Answer: \(a=7\), \(b=14\); \(\mathrm{p}(x)=0\) has only one real root.

(a) Since \(x+2\) is a factor of \(\mathrm p(x)\), the factor theorem gives

\(\mathrm p(-2)=0.\)

Substitute \(x=-2\) into \(\mathrm p(x)=2x^3+ax^2+13x+b\):

\(2(-2)^3+a(-2)^2+13(-2)+b=0.\)

So

\(-16+4a-26+b=0,\)

which gives

\(4a+b=42. \tag{1}\)

The remainder when \(\mathrm p(x)\) is divided by \(x+1\) is \(\mathrm p(-1)\). This remainder is \(6\), so

\(2(-1)^3+a(-1)^2+13(-1)+b=6.\)

Hence

\(-2+a-13+b=6,\)

so

\(a+b=21. \tag{2}\)

Subtracting (2) from (1),

\(3a=21,\)

so \(a=7\). Then from \(a+b=21\),

\(b=14.\)

(b) Therefore

\(\mathrm p(x)=2x^3+7x^2+13x+14.\)

Since \(x+2\) is a factor, divide by \(x+2\):

\(\mathrm p(x)=(x+2)(2x^2+3x+7).\)

Thus \(x=-2\) is one real root. The other possible roots come from

\(2x^2+3x+7=0.\)

The discriminant is

\(3^2-4(2)(7)=9-56=-47.\)

Since this is negative, the quadratic has no real roots. Hence \(\mathrm p(x)=0\) has only one real root.

Answer: \(a=2\), \(b=12\), \(c=-9\).

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.

\(\mathrm p(x)=ax^3+11x^2+bx+c\), so

\(\mathrm p'(x)=3ax^2+22x+b\).

Since \(\mathrm p'(0)=12\), we get \(b=12\).

Since \(x+3\) is a factor, \(\mathrm p(-3)=0\):

\(-27a+99-3b+c=0\).

Using \(b=12\), this becomes

\(-27a+63+c=0\).

The remainder when \(\mathrm p(x)\) is divided by \(x-1\) is \(\mathrm p(1)\), so

\(a+11+b+c=16\).

Using \(b=12\),

\(a+c=-7\).

From \(-27a+63+c=0\), \(c=27a-63\). Substitute into \(a+c=-7\):

\(a+27a-63=-7\), so \(28a=56\), and \(a=2\).

Then \(c=-9\).

This gives the required answer: \(a=2\), \(b=12\), \(c=-9\).

Answer: \(a=6\), \(b=1\), \(c=6\).

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 \(x+2\) is a factor, \(\mathrm p(-2)=0\):

\(-8a+4b+38+c=0.\)

When \(\mathrm p(x)\) is divided by \(x+1\), the remainder is \(\mathrm p(-1)\), so

\(-a+b+19+c=20.\)

Subtracting these equations eliminates \(c\):

\(-8a+4b+38+c)-(-a+b+19+c)=0-20\).

This gives \(-7a+3b+19=-20\), so \(7a-3b=39\), as required.

Now \(\mathrm p'(x)=3ax^2+2bx-19\). Since the remainder when \(\mathrm p'(x)\) is divided by \(x-1\) is \(1\),

\(\mathrm p'(1)=1\).

Thus \(3a+2b-19=1\), so \(3a+2b=20\).

Solve

\(7a-3b=39,\quad 3a+2b=20.\)

This gives \(a=6\), \(b=1\).

Using \(-a+b+19+c=20\),

\(-6+1+19+c=20\), so \(c=6\).

This gives the required answer: \(a=6\), \(b=1\), \(c=6\).

Answer: \(a=6\), \(b=-14\), \(c=-8\), and \(p(x)=(3x-4)(2x+1)(x+2)\).

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) Differentiate twice:

Given \(p''\left(\frac12\right)=32\),

\(6a\left(\frac12\right)+14=32.\)

So

\(3a=18,\)

and therefore

\(a=6.\)

(b) Since \(3x-4\) is a factor, \(x=\frac43\) is a root. With \(a=6\),

\(p(x)=6x^3+7x^2+bx+c.\)

So

\(p\left(\frac43\right)=0.\)

Multiplying by \(9\) after substitution gives

\(128+112+12b+9c=0,\)

so

\(12b+9c=-240.\)

Dividing by \(3\),

\(4b+3c=-80. \tag{1}\)

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

\(p(-1)=7.\)

Thus

\(-6+7-b+c=7,\)

so

\(-b+c=6. \tag{2}\)

From (2), \(c=b+6\). Substitute this into (1):

\(4b+3(b+6)=-80.\)

So

\(7b=-98,\)

giving

(c) Therefore

\(p(x)=6x^3+7x^2-14x-8.\)

Dividing by \(3x-4\) gives

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

(d) Factorise the quadratic:

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

Hence

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

Answer: \(a=6\), \(b=-10\), \(c=0\); \(P(x)=x(3x+2)(2x-5)\).

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 \(P(a)=0\).

Since \(P(x)\) is divisible by \(x\), we have \(P(0)=0\). Therefore

\(c=0\).

The remainder when \(P(x)\) is divided by \(2x+1\) is \(P\left(-\frac12\right)\), so

\(P\left(-\frac12\right)=\frac32\).

Using \(c=0\),

\(-\frac{a}{8}-\frac{11}{4}-\frac{b}{2}=\frac32\).

Multiply by \(-8\):

\(a+4b=-34\).

Now differentiate:

\(P'(x)=3ax^2-22x+b\).

Given \(P'(2)=18\),

\(12a-44+b=18\),

so

\(12a+b=62\).

Solve the simultaneous equations

\(a+4b=-34\)

and

\(12a+b=62\).

From \(a=-34-4b\). Substitute:

\(12(-34-4b)+b=62\).

So \(-408-47b=62\), giving \(b=-10\).

Then \(a+4(-10)=-34\), so \(a=6\).

Thus \(a=6\), \(b=-10\), and \(c=0\).

Now

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

Factorise the quadratic:

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

Therefore

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

Answer: (a) \(a=6\), \(b=7\), \(c=12\); (b) \(p(x)=(3x-1)(2x^2+3x+5)\), so the only real root is \(x=\frac13\).

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) Since

\(p(x)=ax^3+bx^2+cx-5\),

we have

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

Given \(p'(0)=12\), it follows that

\(c=12\).

Since \(3x-1\) is a factor, \(x=\frac13\) is a root. Therefore

\(p\left(\frac13\right)=0\).

Using \(c=12\),

\(\frac{a}{27}+\frac{b}{9}+\frac{12}{3}-5=0\).

So

\(\frac{a}{27}+\frac{b}{9}=1\),

which gives

\(a+3b=27\).

The remainder when divided by \(x-2\) is \(p(2)\). Since the remainder is 95,

\(p(2)=95\).

So

\(8a+4b+2(12)-5=95\).

Hence

\(8a+4b+19=95\),

so

\(2a+b=19\).

Solve

\(a+3b=27\)

and

\(2a+b=19\).

From the second equation, \(b=19-2a\). Substitute into the first:

\(a+3(19-2a)=27\).

So \(a+57-6a=27\), giving \(-5a=-30\), and hence \(a=6\).

Then \(2(6)+b=19\), so \(b=7\).

Therefore \(a=6\), \(b=7\), and \(c=12\).

(b) Now

\(p(x)=6x^3+7x^2+12x-5\).

Since \(3x-1\) is a factor, divide or factorise to get

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

The quadratic factor has discriminant

\(3^2-4(2)(5)=9-40=-31\).

Since the discriminant is negative, \(2x^2+3x+5=0\) has no real roots.

Therefore the only real root of \(p(x)=0\) is from \(3x-1=0\), which gives \(x=\frac13\).

Hence \(p(x)=0\) has only one real root.