Past Exam Questions

(i) To find \(a\) and \(b\), perform polynomial division of \(4x^4 + ax^2 + 11x + b\) by \(x^2 - x + 2\). The division gives a partial quotient of \(4x^2 + kx\).

Obtain the quotient \(4x^2 + 4x + a - 4\) or \(4x^2 + 4x + b/2\).

Equate the \(x\) or constant term to zero and solve for \(a\) or \(b\).

From the mark scheme, \(a = 1\) and \(b = -6\).

(ii) Show that \(x^2 - x + 2 = 0\) has no real roots.

Obtain roots \(\frac{1}{2}\) and \(-\frac{3}{2}\) from \(4x^2 + 4x - 3 = 0\).

(i) Since \((2x + 1)\) is a factor of \(p(x)\), substituting \(x = -\frac{1}{2}\) into \(p(x)\) should yield zero. Thus, \(p\left(-\frac{1}{2}\right) = 4\left(-\frac{1}{2}\right)^3 + a\left(-\frac{1}{2}\right) + 2 = 0\).

Calculating, \(-\frac{1}{2}\) cubed is \(-\frac{1}{8}\), so \(4\left(-\frac{1}{8}\right) = -\frac{1}{2}\).

Thus, \(-\frac{1}{2} - \frac{a}{2} + 2 = 0\).

Solving for \(a\), we get \(-\frac{a}{2} = -\frac{3}{2}\), so \(a = 3\).

(ii)(a) With \(a = 3\), \(p(x) = 4x^3 + 3x + 2\). Performing polynomial division of \(4x^3 + 3x + 2\) by \(2x + 1\), we obtain \(2x^2 - x + 2\) as the quotient.

Thus, \(p(x) = (2x + 1)(2x^2 - x + 2)\).

(ii)(b) The critical value from \(2x + 1 = 0\) is \(x = -\frac{1}{2}\).

The quadratic \(2x^2 - x + 2\) is always positive since its discriminant \((-1)^2 - 4 \times 2 \times 2 = -15\) is negative, indicating no real roots and it opens upwards.

Thus, \(p(x) > 0\) for \(x > -\frac{1}{2}\).

To show that \((x + 1)\) is a factor of \(4x^3 - x^2 - 11x - 6\), we can use the Factor Theorem. According to the theorem, \((x + 1)\) is a factor if substituting \(x = -1\) into the polynomial results in zero.

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

\(4(-1)^3 - (-1)^2 - 11(-1) - 6\)

\(= 4(-1) - 1 + 11 - 6\)

\(= -4 - 1 + 11 - 6\)

\(= 0\)

Since the result is zero, \((x + 1)\) is a factor of \(4x^3 - x^2 - 11x - 6\).

(i) Since \((x + 1)\) is a factor of \(p(x)\), substituting \(x = -1\) into \(p(x)\) gives:

\(-8 + a - b - 1 = 0\)

\(a - b = 9\) (Equation 1)

When \(p(x)\) is divided by \((2x + 1)\), the remainder is 1. Substituting \(x = -\frac{1}{2}\) into \(p(x)\) gives:

\(-1 + \frac{1}{4}a - \frac{1}{2}b - 1 = 1\)

\(\frac{1}{4}a - \frac{1}{2}b = 3\) (Equation 2)

Solving Equation 1 and Equation 2 simultaneously:

From Equation 1: \(a = b + 9\)

Substitute into Equation 2:

\(\frac{1}{4}(b + 9) - \frac{1}{2}b = 3\)

\(\frac{1}{4}b + \frac{9}{4} - \frac{1}{2}b = 3\)

\(-\frac{1}{4}b = 3 - \frac{9}{4}\)

\(-\frac{1}{4}b = \frac{3}{4}\)

\(b = -3\)

Substitute \(b = -3\) into \(a = b + 9\):

\(a = 6\)

(ii) With \(a = 6\) and \(b = -3\), the polynomial is:

\(p(x) = 8x^3 + 6x^2 - 3x - 1\)

Since \((x + 1)\) is a factor, divide \(p(x)\) by \((x + 1)\):

\(8x^3 + 6x^2 - 3x - 1 = (x + 1)(8x^2 - 2x - 1)\)

Factor the quadratic \(8x^2 - 2x - 1\):

\(8x^2 - 2x - 1 = (4x + 1)(2x - 1)\)

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

Since \(p(x)\) is divisible by \((2x - 1)\), substituting \(x = \frac{1}{2}\) into \(p(x)\) gives:

\(\frac{1}{4} + \frac{a}{4} - \frac{11}{2} + b = 0\)

Multiplying through by 4 to clear fractions:

\(1 + a - 22 + 4b = 0\)

\(a + 4b = 21\)

When \(p(x)\) is divided by \((x + 1)\), the remainder is 12. Substituting \(x = -1\) into \(p(x)\) gives:

\(-2 + a + 11 + b = 12\)

\(a + b = 3\)

Solving the system of equations:

1. \(a + 4b = 21\)

2. \(a + b = 3\)

Subtract equation 2 from equation 1:

\((a + 4b) - (a + b) = 21 - 3\)

\(3b = 18\)

\(b = 6\)

Substitute \(b = 6\) into \(a + b = 3\):

\(a + 6 = 3\)

\(a = -3\)

Thus, \(a = -3\) and \(b = 6\).

(i) Since \((x+1)\) and \((x+2)\) are factors of \(p(x)\), we have:

\(p(-1) = 0\) and \(p(-2) = 0\).

Substituting \(x = -1\) into \(p(x)\):

\(4(-1)^3 + a(-1)^2 + b(-1) - 2 = 0\)

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

\(a - b = 6\) (Equation 1)

Substituting \(x = -2\) into \(p(x)\):

\(4(-2)^3 + a(-2)^2 + b(-2) - 2 = 0\)

\(-32 + 4a - 2b - 2 = 0\)

\(4a - 2b = 34\) (Equation 2)

Solving Equation 1 and Equation 2 simultaneously:

From Equation 1: \(a = b + 6\)

Substitute into Equation 2:

\(4(b + 6) - 2b = 34\)

\(4b + 24 - 2b = 34\)

\(2b = 10\)

\(b = 5\)

Substitute \(b = 5\) into \(a = b + 6\):

\(a = 11\)

(ii) With \(a = 11\) and \(b = 5\), the polynomial is:

\(p(x) = 4x^3 + 11x^2 + 5x - 2\)

We need to find the remainder when \(p(x)\) is divided by \((x^2 + 1)\).

Using polynomial division or synthetic division, the quotient is \(4x + a\) and the remainder is \(x - 13\).

Since \((3x + 1)\) is a factor of \(p(x)\), substituting \(x = -\frac{1}{3}\) into \(p(x)\) should yield zero:

\(p\left(-\frac{1}{3}\right) = a\left(-\frac{1}{3}\right)^3 + b\left(-\frac{1}{3}\right)^2 + \left(-\frac{1}{3}\right) + 3 = 0\)

\(-\frac{1}{27}a + \frac{1}{9}b - \frac{1}{3} + 3 = 0\)

\(-\frac{1}{27}a + \frac{1}{9}b + \frac{8}{3} = 0\)

Multiply through by 27 to clear fractions:

\(-a + 3b + 72 = 0\)

\(a = 3b + 72\) (Equation 1)

When \(p(x)\) is divided by \((x - 2)\), the remainder is 21. Substitute \(x = 2\) into \(p(x)\):

\(p(2) = a(2)^3 + b(2)^2 + 2 + 3 = 21\)

\(8a + 4b + 5 = 21\)

\(8a + 4b = 16\)

\(2a + b = 4\) (Equation 2)

Substitute \(a = 3b + 72\) from Equation 1 into Equation 2:

\(2(3b + 72) + b = 4\)

\(6b + 144 + b = 4\)

\(7b + 144 = 4\)

\(7b = -140\)

\(b = -20\)

Substitute \(b = -20\) back into Equation 1:

\(a = 3(-20) + 72\)

\(a = -60 + 72\)

\(a = 12\)

Thus, the values are \(a = 12\) and \(b = -20\).

(i) Since \((x + 2)\) is a factor of \(f(x)\), substituting \(x = -2\) into \(f(x)\) should yield zero:

\(f(-2) = (-2)^3 + a(-2)^2 - a(-2) + 14 = 0\)

\(-8 + 4a + 2a + 14 = 0\)

\(6a + 6 = 0\)

\(6a = -6\)

\(a = -1\)

(ii) With \(a = -1\), the polynomial becomes:

\(f(x) = x^3 - x^2 + x + 14\)

We need to show that \(f(x) = 0\) has only one real root. First, find the quadratic factor by dividing \(f(x)\) by \((x + 2)\):

\(f(x) = (x + 2)(x^2 - 3x + 7)\)

Now, use the discriminant of the quadratic \(x^2 - 3x + 7\):

\(b^2 - 4ac = (-3)^2 - 4(1)(7) = 9 - 28 = -19\)

Since the discriminant is negative, \(x^2 - 3x + 7 = 0\) has no real roots. Therefore, \(f(x) = 0\) has only one real root, which is \(x = -2\).

(i) Since \((2x + 1)\) is a factor of \(p(x)\), substituting \(x = -\frac{1}{2}\) into \(p(x)\) gives:

\(8\left(-\frac{1}{2}\right)^3 + a\left(-\frac{1}{2}\right)^2 + b\left(-\frac{1}{2}\right) + 3 = 0\)

Simplifying, we get:

\(-1 + \frac{a}{4} - \frac{b}{2} + 3 = 0\)

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

When \(p(x)\) is divided by \((2x - 1)\), substituting \(x = \frac{1}{2}\) gives a remainder of 1:

\(8\left(\frac{1}{2}\right)^3 + a\left(\frac{1}{2}\right)^2 + b\left(\frac{1}{2}\right) + 3 = 1\)

Simplifying, we get:

\(1 + \frac{a}{4} + \frac{b}{2} + 3 = 1\)

\(a + 2b + 12 = 0\)

Solving the equations \(a - 2b + 8 = 0\) and \(a + 2b + 12 = 0\), we find:

\(a = -10\), \(b = -1\)

(ii) With \(a = -10\) and \(b = -1\), divide \(p(x) = 8x^3 - 10x^2 - x + 3\) by \(2x^2 - 1\):

The division gives a quotient of \(4x - 5\) and a remainder of \(3x - 2\).

(i) Since \((3x + 1)\) is a factor of \(p(x)\), substituting \(x = -\frac{1}{3}\) into \(p(x)\) should yield zero:

\(a\left(-\frac{1}{3}\right)^3 - 20\left(-\frac{1}{3}\right)^2 + \left(-\frac{1}{3}\right) + 3 = 0\)

\(-\frac{a}{27} - \frac{20}{9} - \frac{1}{3} + 3 = 0\)

Solving for \(a\):

\(-\frac{a}{27} = \frac{20}{9} + \frac{1}{3} - 3\)

\(-\frac{a}{27} = \frac{20}{9} + \frac{3}{9} - \frac{27}{9}\)

\(-\frac{a}{27} = -\frac{4}{9}\)

\(a = 12\)

(ii) With \(a = 12\), the polynomial becomes \(12x^3 - 20x^2 + x + 3\). Divide by \(3x + 1\) to find the quadratic factor:

Perform polynomial division to get \(4x^2 - 8x + 3\).

Factor the quadratic:

\(4x^2 - 8x + 3 = (2x - 1)(2x - 3)\)

Thus, the complete factorisation is \((3x+1)(2x-1)(2x-3)\).

To divide \(2x^2\) by \(x + 2\), we perform polynomial long division.

1. Divide the first term of the dividend \(2x^2\) by the first term of the divisor \(x\):

\(\frac{2x^2}{x} = 2x\)

2. Multiply the entire divisor \(x + 2\) by \(2x\):

\(2x(x + 2) = 2x^2 + 4x\)

3. Subtract \(2x^2 + 4x\) from \(2x^2\):

\(2x^2 - (2x^2 + 4x) = -4x\)

4. Divide \(-4x\) by \(x\):

\(\frac{-4x}{x} = -4\)

5. Multiply the entire divisor \(x + 2\) by \(-4\):

\(-4(x + 2) = -4x - 8\)

6. Subtract \(-4x - 8\) from \(-4x\):

\(-4x - (-4x - 8) = 8\)

The quotient is \(2x - 4\) and the remainder is 8.

(i) Since \((x - 2)\) is a factor of \(p(x)\), by the factor theorem, \(p(2) = 0\).

Substitute \(x = 2\) into \(p(x)\):

\(p(2) = 2^3 - 3a(2) + 4a = 0\)

\(8 - 6a + 4a = 0\)

\(8 - 2a = 0\)

\(2a = 8\)

\(a = 4\)

(ii)(a) Substitute \(a = 4\) into \(p(x)\):

\(p(x) = x^3 - 12x + 16\)

Divide \(p(x)\) by \((x-2)\) to find the other factor:

\(x^3 - 12x + 16 = (x-2)(x^2 + 2x - 8)\)

Factor \(x^2 + 2x - 8\):

\(x^2 + 2x - 8 = (x-2)(x+4)\)

Thus, \(p(x) = (x-2)^2(x+4)\)

(ii)(b) Substitute \(x^2\) into \(p(x)\):

\(p(x^2) = (x^2 - 2)^2(x^2 + 4)\)

Set \(p(x^2) = 0\):

\((x^2 - 2)^2 = 0\) gives \(x^2 = 2\) so \(x = \pm \sqrt{2}\)

\((x^2 + 4) = 0\) gives \(x^2 = -4\) so \(x = \pm 2i\)

The roots are \(\pm \sqrt{2}, \pm 2i\).

Since \((2x - 1)\) is a factor of \(p(x)\), substituting \(x = \frac{1}{2}\) into \(p(x)\) should yield zero:

\(p\left(\frac{1}{2}\right) = a\left(\frac{1}{2}\right)^3 - \left(\frac{1}{2}\right)^2 + 4\left(\frac{1}{2}\right) - a = 0\)

\(\Rightarrow \frac{a}{8} - \frac{1}{4} + 2 - a = 0\)

\(\Rightarrow \frac{a}{8} - a = -\frac{7}{4}\)

\(\Rightarrow \frac{a - 8a}{8} = -\frac{7}{4}\)

\(\Rightarrow -\frac{7a}{8} = -\frac{7}{4}\)

\(\Rightarrow 7a = 14\)

\(\Rightarrow a = 2\)

Substituting \(a = 2\) back into \(p(x)\):

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

We know \((2x - 1)\) is a factor, so divide \(p(x)\) by \((2x - 1)\) to find the other factor:

Using polynomial division or inspection, we find:

\(p(x) = (2x - 1)(x^2 + 2)\)

(i) To find the value of \(a\), we divide \(p(x) = x^4 + 3x^3 + ax + 3\) by \(x^2 - x + 1\). Performing polynomial division, we obtain a quotient of \(x^2 + 4x + 3\) and a remainder of zero, which confirms divisibility. Solving for \(a\), we find \(a = 1\).

(ii) With \(a = 1\), the polynomial becomes \(x^4 + 3x^3 + x + 3\). Factoring or solving this equation, we find the real roots are \(x = -3\) and \(x = -1\).

(i) Since \((2x - 1)\) is a factor of \(p(x)\), substituting \(x = \frac{1}{2}\) into \(p(x)\) gives:

\(\frac{1}{8}a + \frac{1}{4}b + \frac{5}{2} - 2 = 0\)

\(\Rightarrow \frac{1}{8}a + \frac{1}{4}b + \frac{1}{2} = 0\)

\(\Rightarrow a + 2b + 4 = 0\) (Equation 1)

When \(p(x)\) is divided by \((x - 2)\), the remainder is 12. Substituting \(x = 2\) into \(p(x)\) gives:

\(8a + 4b + 10 - 2 = 12\)

\(\Rightarrow 8a + 4b + 8 = 12\)

\(\Rightarrow 8a + 4b = 4\) (Equation 2)

Solving Equations 1 and 2 simultaneously:

From Equation 1: \(a = -2b - 4\)

Substitute into Equation 2:

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

\(-16b - 32 + 4b = 4\)

\(-12b = 36\)

\(b = -3\)

Substitute \(b = -3\) into Equation 1:

\(a + 2(-3) + 4 = 0\)

\(a - 6 + 4 = 0\)

\(a = 2\)

(ii) With \(a = 2\) and \(b = -3\), the polynomial is \(2x^3 - 3x^2 + 5x - 2\).

Divide \(2x^3 - 3x^2 + 5x - 2\) by \(2x - 1\) to find the quadratic factor:

The division gives a quotient of \(x^2 - x + 2\).

Thus, the quadratic factor is \(x^2 - x + 2\).

To divide \(2x^4 - 27\) by \(x^2 + x + 3\), we perform polynomial long division:

1. Divide the leading term \(2x^4\) by \(x^2\) to get \(2x^2\).

2. Multiply \(2x^2\) by \(x^2 + x + 3\) to get \(2x^4 + 2x^3 + 6x^2\).

3. Subtract \(2x^4 + 2x^3 + 6x^2\) from \(2x^4 - 27\) to get \(-2x^3 - 6x^2 - 27\).

4. Divide \(-2x^3\) by \(x^2\) to get \(-2x\).

5. Multiply \(-2x\) by \(x^2 + x + 3\) to get \(-2x^3 - 2x^2 - 6x\).

6. Subtract \(-2x^3 - 2x^2 - 6x\) from \(-2x^3 - 6x^2 - 27\) to get \(-4x^2 - 6x - 27\).

7. Divide \(-4x^2\) by \(x^2\) to get \(-4\).

8. Multiply \(-4\) by \(x^2 + x + 3\) to get \(-4x^2 - 4x - 12\).

9. Subtract \(-4x^2 - 4x - 12\) from \(-4x^2 - 6x - 27\) to get \(-2x - 15\).

The quotient is \(2x^2 - 2x - 4\) and the remainder is \(10x - 15\).

(i) To show that \(f(-2) = 0\), substitute \(x = -2\) into \(f(x)\):

\(f(-2) = 12(-2)^3 + 25(-2)^2 - 4(-2) - 12\)

\(= -96 + 100 + 8 - 12\)

\(= 0\)

Thus, \(x + 2\) is a factor of \(f(x)\).

To factorise \(f(x)\) completely, divide \(f(x)\) by \(x + 2\):

Using synthetic division or polynomial division, we find:

\(12x^3 + 25x^2 - 4x - 12 = (x+2)(12x^2 + x - 6)\)

Factor the quadratic \(12x^2 + x - 6\):

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

Thus, \(f(x) = (x+2)(4x+3)(3x-2)\).

(ii) Given \(12 \times 27^y + 25 \times 9^y - 4 \times 3^y - 12 = 0\), let \(3^y = k\).

Then \(9^y = k^2\) and \(27^y = k^3\).

Substitute into the equation:

\(12k^3 + 25k^2 - 4k - 12 = 0\)

By inspection or trial, \(k = \frac{2}{3}\) satisfies the equation.

Thus, \(3^y = \frac{2}{3}\).

Taking logarithms, \(y = \log_3 \left( \frac{2}{3} \right)\).

Calculate \(y\) to 3 significant figures:

\(y \approx -0.369\).

(i) Since \((z + 2)\) is a factor of \(p(z)\), we have \(p(-2) = 0\).

Substitute \(z = -2\) into \(p(z)\):

\((-2)^3 + m(-2)^2 + 24(-2) + 32 = 0\)

\(-8 + 4m - 48 + 32 = 0\)

\(4m - 24 = 0\)

\(4m = 24\)

\(m = 6\)

(ii)(a) With \(m = 6\), the polynomial becomes \(p(z) = z^3 + 6z^2 + 24z + 32\).

Since \(z = -2\) is a root, divide \(p(z)\) by \((z + 2)\) to find the quadratic factor:

\(z^3 + 6z^2 + 24z + 32 = (z + 2)(z^2 + 4z + 16)\)

Solving \(z^2 + 4z + 16 = 0\) using the quadratic formula:

\(z = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot 16}}{2 \cdot 1}\)

\(z = \frac{-4 \pm \sqrt{-48}}{2}\)

\(z = \frac{-4 \pm 4\sqrt{3}i}{2}\)

\(z = -2 \pm 2\sqrt{3}i\)

Thus, the roots are \(z = -2, -2 + 2\sqrt{3}i, -2 - 2\sqrt{3}i\).

(ii)(b) To find the roots of \(p(z^2) = 0\), substitute \(z^2\) into the roots found in part (ii)(a):

For \(z = -2\), \(z^2 = 4\), so \(z = \pm i\sqrt{2}\).

For \(z = -2 + 2\sqrt{3}i\), find the square roots:

Let \(z = a + bi\), then \(a^2 - b^2 = -2\) and \(2ab = 2\sqrt{3}\).

Solving gives \(a = 1, b = \sqrt{3}\) or \(a = -1, b = -\sqrt{3}\).

Thus, \(z = 1 + i\sqrt{3}, 1 - i\sqrt{3}, -1 + i\sqrt{3}, -1 - i\sqrt{3}\).

The six roots of \(p(z^2) = 0\) are \(\pm i\sqrt{2}, \pm (1+i\sqrt{3}), \pm (1-i\sqrt{3})\).

(i) Since \((2x + 1)\) is a factor of \(p(x)\), substituting \(x = -\frac{1}{2}\) into \(p(x)\) gives:

\(-\frac{1}{4} + \frac{5}{4} - \frac{1}{2}a + b = 0\)

\(\Rightarrow -\frac{1}{2}a + b = -1\)

When \(x = -2\), the remainder is 9:

\(-16 + 20 - 2a + b = 9\)

\(\Rightarrow -2a + b = 5\)

Solving the equations:

\(-\frac{1}{2}a + b = -1\)

\(-2a + b = 5\)

Subtract the first from the second:

\(-2a + b + \frac{1}{2}a - b = 5 + 1\)

\(-\frac{3}{2}a = 6\)

\(a = -4\)

Substitute \(a = -4\) into \(-\frac{1}{2}a + b = -1\):

\(2 + b = -1\)

\(b = -3\)

(ii) With \(a = -4\) and \(b = -3\), \(p(x) = 2x^3 + 5x^2 - 4x - 3\).

Divide \(p(x)\) by \(2x + 1\) to get a quotient of \(x^2 + 2x - 3\).

Factorise \(x^2 + 2x - 3\) as \((x + 3)(x - 1)\).

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

(i) To find the value of \(a\), we perform polynomial division of \(4x^3 - 4x^2 + 3x + a\) by \(2x^2 - 3x + 3\). The division should result in a remainder of zero since \(p(x)\) is divisible by \(2x^2 - 3x + 3\).

Performing the division, we find that the remainder is zero when \(a = 3\).

(ii) With \(a = 3\), the polynomial becomes \(4x^3 - 4x^2 + 3x + 3\). We need to solve the inequality \(p(x) < 0\).

Since \(2x^2 - 3x + 3\) is never zero and always positive, we analyze the sign of \(p(x)\) by considering the roots of the polynomial.

The inequality \(p(x) < 0\) holds when \(x < -\frac{1}{2}\), as justified by the behavior of the polynomial and its derivative.