Past Exam Questions

Since \(x^2 + x + 2\) is a factor of \(p(x) = x^4 + 3x^2 + a\), we can perform polynomial division of \(p(x)\) by \(x^2 + x + 2\).

1. Divide \(x^4 + 3x^2 + a\) by \(x^2 + x + 2\) to get a quotient of \(x^2 - x + 2\).

2. The remainder must be zero for \(x^2 + x + 2\) to be a factor.

3. Equate the constant remainder to zero and solve for \(a\):

\(a = 4\).

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

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

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

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

\(a = 4\)

(ii) With \(a = 4\), the polynomial becomes \(x^3 - 2x + 4\). To find the quadratic factor, divide \(x^3 - 2x + 4\) by \(x + 2\):

Perform polynomial division:

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

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

(i) To find \(a\), perform polynomial division of \(x^4 + 5x + a\) by \(x^2 - x + 3\). The quotient is \(x^2 + x\) and the remainder is \(3x + a\). Equating the remainder to zero gives:

\(3x + a = 0\)

Since \(x^2 - x + 3\) is a factor, the remainder must be zero for all \(x\). Thus, \(a = -6\).

Now, factorise \(x^2 + x - 2\) as \((x + 2)(x - 1)\). Therefore, \(p(x) = (x^2 - x + 3)(x + 2)(x - 1)\).

(ii) The quadratic \(x^2 + x - 2 = 0\) has two real roots, \(x = -2\) and \(x = 1\). The quadratic \(x^2 - x + 3 = 0\) has no real roots (discriminant \(b^2 - 4ac = 1 - 12 = -11\)). Therefore, \(p(x) = 0\) has two real roots.

(i) Since \((x - 2)\) is a factor of \(p(x)\), substituting \(x = 2\) into \(p(x)\) should yield zero: \(2(2)^3 + a(2)^2 - 4 = 0\). Simplifying gives \(16 + 4a - 4 = 0\), leading to \(4a = -12\), so \(a = -3\).

(ii) With \(a = -3\), \(p(x) = 2x^3 - 3x^2 - 4\). Dividing by \((x - 2)\) gives the quadratic factor \(2x^2 + x + 2\). Thus, \(p(x) = (x-2)(2x^2 + x + 2)\).

(iii) To solve \(p(x) > 0\), note that \(2x^2 + x + 2\) has no real roots and is always positive. Therefore, \(p(x) > 0\) when \(x > 2\).

(i) Since \(f(x)\) is divisible by \(x^2 - 4x + 4\), we can write \(f(x) = (x^2 - 4x + 4)(x^2 + bx + c)\).

By comparing coefficients, we find that \(b = 2\) and \(4c = a\). Solving for \(a\), we get \(a = 8\).

(ii) Substitute \(a = 8\) into \(f(x)\), giving \(f(x) = (x^2 - 4x + 4)(x^2 + 2x + 2)\).

Since \(x^2 - 4x + 4 = (x-2)^2\), it is always non-negative.

For \(x^2 + 2x + 2\), the discriminant is \(2^2 - 4 \times 1 \times 2 = -4\), which is negative, indicating no real roots and always positive.

Thus, \(f(x)\) is the product of two non-negative expressions, making \(f(x)\) never negative.

Given that \((x^2 + x + 2)\) is a factor of \(p(x) = x^4 + 4x^2 + x + a\), we can express \(p(x)\) as:

\(p(x) = (x^2 + x + 2)(x^2 + bx + c)\)

Expanding the right-hand side, we have:

\((x^2 + x + 2)(x^2 + bx + c) = x^4 + bx^3 + cx^2 + x^3 + bx^2 + cx + 2x^2 + 2bx + 2c\)

Combining like terms, we get:

\(x^4 + (b+1)x^3 + (c+b+2)x^2 + (c+2b)x + 2c\)

Equating coefficients with \(x^4 + 4x^2 + x + a\), we have:

1. \(b+1 = 0\) implies \(b = -1\)

2. \(c+b+2 = 4\) implies \(c - 1 + 2 = 4\) so \(c = 3\)

3. \(c+2b = 1\) implies \(3 - 2 = 1\), which is consistent

4. \(2c = a\) implies \(2 \times 3 = a\) so \(a = 6\)

Thus, the other quadratic factor is \(x^2 - x + 3\).

To find the values of \(a\) and \(b\), we perform polynomial division of \(2x^4 + ax^3 + bx - 1\) by \(x^2 - x + 1\) and set the remainder equal to \(3x + 2\).

1. Start the division: The first term of the quotient is \(2x^2\) because \(2x^2 \cdot (x^2 - x + 1) = 2x^4 - 2x^3 + 2x^2\).

2. Subtract \(2x^4 - 2x^3 + 2x^2\) from \(2x^4 + ax^3 + bx - 1\) to get \((a + 2)x^3 - 2x^2 + bx - 1\).

3. The next term of the quotient is \((a + 2)x\) because \((a + 2)x \cdot (x^2 - x + 1) = (a + 2)x^3 - (a + 2)x^2 + (a + 2)x\).

4. Subtract \((a + 2)x^3 - (a + 2)x^2 + (a + 2)x\) from \((a + 2)x^3 - 2x^2 + bx - 1\) to get \((b - (a + 2))x - 1\).

5. The remainder is \((b - (a + 2))x - 1\), which must equal \(3x + 2\).

6. Equate coefficients: \(b - (a + 2) = 3\) and \(-1 = 2\).

7. Solve the equations: From \(-1 = 2\), we find \(a = -3\). From \(b - (a + 2) = 3\), substitute \(a = -3\) to get \(b - (-3 + 2) = 3\), which simplifies to \(b + 1 = 3\), so \(b = 5\).

Thus, \(a = -3\) and \(b = 5\).

Since \(p(x)\) is divisible by \((2x - 1)\), we have \(p\left(\frac{1}{2}\right) = 0\).

Substitute \(x = \frac{1}{2}\) into \(p(x)\):

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

\(\frac{1}{8}a + \frac{1}{4} + \frac{1}{2}b + 3 = 0\)

Multiply through by 8 to clear fractions:

\(a + 2 + 4b + 24 = 0\)

\(a + 4b = -26\)

When \(p(x)\) is divided by \((x + 2)\), the remainder is 5, so \(p(-2) = 5\).

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

\(a(-2)^3 + (-2)^2 + b(-2) + 3 = 5\)

\(-8a + 4 - 2b + 3 = 5\)

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

\(8a + 2b = 2\)

Now solve the system of equations:

1. \(a + 4b = -26\)

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

Multiply equation 1 by 2:

\(2a + 8b = -52\)

Subtract equation 2 from this result:

\(2a + 8b - 8a - 2b = -52 - 2\)

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

\(-a + b = -9\)

\(b = a - 9\)

Substitute \(b = a - 9\) into equation 1:

\(a + 4(a - 9) = -26\)

\(a + 4a - 36 = -26\)

\(5a = 10\)

\(a = 2\)

Substitute \(a = 2\) back to find \(b\):

\(b = 2 - 9 = -7\)

Thus, \(a = 2\) and \(b = -7\).

(a) Since \((x-2)\) is a factor of \(p(x)\), substituting \(x = 2\) into \(p(x)\) gives:

\(8a - 40 + 2b + 8 = 0\)

\(8a + 2b - 32 = 0\)

\(4a + b = 16\) (Equation 1)

Since \((x-2)\) is also a factor of \(p'(x)\), differentiate \(p(x)\):

\(p'(x) = 3ax^2 - 20x + b\)

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

\(12a - 40 + b = 0\)

\(12a + b = 40\) (Equation 2)

Subtract Equation 1 from Equation 2:

\(12a + b - (4a + b) = 40 - 16\)

\(8a = 24\)

\(a = 3\)

Substitute \(a = 3\) into Equation 1:

\(4(3) + b = 16\)

\(12 + b = 16\)

\(b = 4\)

(b) With \(a = 3\) and \(b = 4\), \(p(x) = 3x^3 - 10x^2 + 4x + 8\).

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

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

Factorise \(3x^2 - 4x - 4\):

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

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

Final factorisation: \(3(x + \frac{2}{3})(x-2)^2\)

To divide \(8x^3 + 4x^2 + 2x + 7\) by \(4x^2 + 1\), we perform polynomial long division.

1. Divide the leading term \(8x^3\) by \(4x^2\) to get \(2x\).

2. Multiply \(2x\) by \(4x^2 + 1\) to get \(8x^3 + 2x\).

3. Subtract \(8x^3 + 2x\) from \(8x^3 + 4x^2 + 2x + 7\) to get \(4x^2 + 7\).

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

5. Multiply \(1\) by \(4x^2 + 1\) to get \(4x^2 + 1\).

6. Subtract \(4x^2 + 1\) from \(4x^2 + 7\) to get \(6\).

The quotient is \(2x + 1\) and the remainder is \(6\).

To divide \(2x^4 + 1\) by \(x^2 - x + 2\), we perform polynomial long division.

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

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

3. Subtract \(2x^4 - 2x^3 + 4x^2\) from \(2x^4 + 1\) to get \(2x^3 - 4x^2 + 1\).

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

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

6. Subtract \(2x^3 - 2x^2 + 4x\) from \(2x^3 - 4x^2 + 1\) to get \(-2x^2 - 4x + 1\).

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

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

9. Subtract \(-2x^2 + 2x - 4\) from \(-2x^2 - 4x + 1\) to get \(-6x + 5\).

The quotient is \(2x^2 + 2x - 2\) and the remainder is \(-6x + 5\).