Lesson 2 of 8
25% complete
1. Quadratic Equations — The Discriminant and Nature of Roots
Quadratic Equation: ax² + bx + c = 0 (a ≠ 0). The discriminant Δ = b² − 4ac determines the nature and number of real roots without solving the equation.
The Discriminant — Nature of Roots
Δ = b² − 4ac
Δ > 0 → two distinct real roots (graph crosses x-axis twice)
Δ = 0 → one repeated (equal) real root (graph touches x-axis once — tangent to x-axis)
Δ < 0 → no real roots (graph does not intersect x-axis)
📐 Worked Example 1 — Using the Discriminant
Find the values of k for which kx² + (k+2)x + 1 = 0 has (a) two distinct real roots (b) equal roots (c) no real roots.
1
Identify: a = k, b = k+2, c = 1. Calculate Δ:
Δ = (k+2)² − 4(k)(1) = k²+4k+4 − 4k = k²+4
Wait — let me re-examine: Δ = (k+2)² − 4k = k²+4k+4−4k = k²+4.
Hmm, k²+4 > 0 for all real k. So Δ is always positive unless k=0 (not quadratic).
Δ = (k+2)² − 4(k)(1) = k²+4k+4 − 4k = k²+4
Wait — let me re-examine: Δ = (k+2)² − 4k = k²+4k+4−4k = k²+4.
Hmm, k²+4 > 0 for all real k. So Δ is always positive unless k=0 (not quadratic).
2
Use a better example: kx² − 3x + k = 0.
a=k, b=−3, c=k. Δ = 9 − 4k²
a=k, b=−3, c=k. Δ = 9 − 4k²
3
(a) Two distinct real roots: Δ > 0
9 − 4k² > 0 → 4k² < 9 → k² < 9/4 → −3/2 < k < 3/2 (and k ≠ 0)
9 − 4k² > 0 → 4k² < 9 → k² < 9/4 → −3/2 < k < 3/2 (and k ≠ 0)
4
(b) Equal roots: Δ = 0
9 − 4k² = 0 → k² = 9/4 → k = ±3/2
9 − 4k² = 0 → k² = 9/4 → k = ±3/2
5
(c) No real roots: Δ < 0
9 − 4k² < 0 → k < −3/2 or k > 3/2
9 − 4k² < 0 → k < −3/2 or k > 3/2
Quadratic Inequalities
Method for Solving Quadratic Inequalities:
1. Rearrange to get 0 on one side: f(x) > 0, f(x) < 0, f(x) ≥ 0, or f(x) ≤ 0
2. Factorise (or find roots using the quadratic formula)
3. Sketch the parabola — identify where it is above/below the x-axis
4. Write the solution using inequalities
Key pattern:
f(x) = a(x−p)(x−q) with a > 0 (U-shape):
f(x) < 0 → p < x < q (BETWEEN the roots — single interval)
f(x) > 0 → x < p or x > q (OUTSIDE the roots — two separate regions)
1. Rearrange to get 0 on one side: f(x) > 0, f(x) < 0, f(x) ≥ 0, or f(x) ≤ 0
2. Factorise (or find roots using the quadratic formula)
3. Sketch the parabola — identify where it is above/below the x-axis
4. Write the solution using inequalities
Key pattern:
f(x) = a(x−p)(x−q) with a > 0 (U-shape):
f(x) < 0 → p < x < q (BETWEEN the roots — single interval)
f(x) > 0 → x < p or x > q (OUTSIDE the roots — two separate regions)
📐 Worked Example 2 — Quadratic Inequalities
Solve: (a) x² − 5x + 6 < 0 (b) 2x² + x − 6 ≥ 0 (c) x² − 6x + 9 > 0
1
(a) Factorise: (x−2)(x−3) < 0
Roots: x=2 and x=3. U-shape parabola is below x-axis BETWEEN roots.
2 < x < 3
Roots: x=2 and x=3. U-shape parabola is below x-axis BETWEEN roots.
2 < x < 3
2
(b) Factorise: (2x−3)(x+2) ≥ 0
Roots: x=3/2 and x=−2. U-shape is above x-axis OUTSIDE roots.
x ≤ −2 or x ≥ 3/2
Roots: x=3/2 and x=−2. U-shape is above x-axis OUTSIDE roots.
x ≤ −2 or x ≥ 3/2
3
(c) x²−6x+9 = (x−3)². This is a perfect square ≥ 0 always.
(x−3)² > 0 when x ≠ 3 (equals 0 only at x=3).
x ∈ ℝ, x ≠ 3
(x−3)² > 0 when x ≠ 3 (equals 0 only at x=3).
x ∈ ℝ, x ≠ 3
Completing the Square — Advanced Applications
📐 Worked Example 3 — Maximum/Minimum Values
Express f(x) = 3x² − 12x + 7 in the form a(x+p)²+q. Hence state: (a) the minimum value of f(x) (b) the value of x at which it occurs (c) the range of f (d) the equation of the axis of symmetry.
1
Factor out 3 from the first two terms:
3(x² − 4x) + 7
3(x² − 4x) + 7
2
Complete the square inside: half of (−4) = −2, square = 4:
3(x² − 4x + 4 − 4) + 7 = 3[(x−2)² − 4] + 7
= 3(x−2)² − 12 + 7 = 3(x−2)² − 5
3(x² − 4x + 4 − 4) + 7 = 3[(x−2)² − 4] + 7
= 3(x−2)² − 12 + 7 = 3(x−2)² − 5
3
(a) Minimum value = −5 (since 3(x−2)² ≥ 0, minimum when (x−2)²=0)
(b) Occurs at x = 2
(c) Range: f(x) ≥ −5
(d) Axis of symmetry: x = 2
(b) Occurs at x = 2
(c) Range: f(x) ≥ −5
(d) Axis of symmetry: x = 2
Condition for ax² + bx + c Always Positive or Always Negative
For quadratic ax² + bx + c (a ≠ 0):
Always positive (above x-axis for all x): a > 0 AND Δ < 0
Always negative (below x-axis for all x): a < 0 AND Δ < 0
Always non-negative (touches x-axis): a > 0 AND Δ = 0
This is extremely useful for proving inequalities and finding ranges of expressions.
Always positive (above x-axis for all x): a > 0 AND Δ < 0
Always negative (below x-axis for all x): a < 0 AND Δ < 0
Always non-negative (touches x-axis): a > 0 AND Δ = 0
This is extremely useful for proving inequalities and finding ranges of expressions.
📐 Worked Example 4 — Always Positive
Show that x² + 3x + 5 > 0 for all real x.
1
Check a: a = 1 > 0 ✓ (U-shaped parabola)
2
Calculate Δ: Δ = 9 − 4(1)(5) = 9 − 20 = −11 < 0 ✓
3
Since a > 0 and Δ < 0, the parabola opens upward and never crosses the x-axis.
Therefore x² + 3x + 5 > 0 for all x ∈ ℝ. ✓
Therefore x² + 3x + 5 > 0 for all x ∈ ℝ. ✓
2. Polynomials — Division, Remainder Theorem, Factor Theorem
Polynomial: An expression of the form aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₁x + a₀ where n is a non-negative integer and aₙ ≠ 0. The degree is the highest power of x.
Polynomial Long Division
When dividing polynomial f(x) by divisor d(x):
f(x) = d(x) × q(x) + r(x)
where q(x) is the quotient and r(x) is the remainder. The degree of r(x) < degree of d(x).
When dividing by a linear factor (x−a), the remainder is a constant.
f(x) = d(x) × q(x) + r(x)
where q(x) is the quotient and r(x) is the remainder. The degree of r(x) < degree of d(x).
When dividing by a linear factor (x−a), the remainder is a constant.
📐 Worked Example 5 — Polynomial Long Division
Divide f(x) = 2x³ + 3x² − 8x + 3 by (x − 1). Hence fully factorise f(x).
1
Set up long division: Divide 2x³ by x → 2x².
2x²(x−1) = 2x³−2x². Remainder: (2x³+3x²) − (2x³−2x²) = 5x²
2x²(x−1) = 2x³−2x². Remainder: (2x³+3x²) − (2x³−2x²) = 5x²
2
Bring down: 5x²−8x. Divide 5x² by x → 5x.
5x(x−1) = 5x²−5x. Remainder: (5x²−8x) − (5x²−5x) = −3x
5x(x−1) = 5x²−5x. Remainder: (5x²−8x) − (5x²−5x) = −3x
3
Bring down: −3x+3. Divide −3x by x → −3.
−3(x−1) = −3x+3. Remainder: (−3x+3)−(−3x+3) = 0
−3(x−1) = −3x+3. Remainder: (−3x+3)−(−3x+3) = 0
4
Quotient = 2x²+5x−3. Remainder = 0, confirming (x−1) is a factor.
Factorise 2x²+5x−3 = (2x−1)(x+3)
f(x) = (x−1)(2x−1)(x+3)
Factorise 2x²+5x−3 = (2x−1)(x+3)
f(x) = (x−1)(2x−1)(x+3)
The Remainder Theorem
Remainder Theorem
When f(x) is divided by (x − a), the remainder = f(a)
When f(x) is divided by (ax − b), the remainder = f(b/a)
This allows finding the remainder WITHOUT performing long division.
The Factor Theorem
Factor Theorem
(x − a) is a factor of f(x) ⟺ f(a) = 0
(ax − b) is a factor of f(x) ⟺ f(b/a) = 0
The factor theorem is a special case of the remainder theorem where the remainder is zero.
📐 Worked Example 6 — Remainder and Factor Theorems
f(x) = x³ + ax² − 7x + b. When f(x) is divided by (x−2) the remainder is 6. (x+3) is a factor of f(x). Find a and b, then fully factorise f(x).
1
Using Remainder Theorem: f(2) = 6
8 + 4a − 14 + b = 6 → 4a + b = 12 ... (1)
8 + 4a − 14 + b = 6 → 4a + b = 12 ... (1)
2
Using Factor Theorem: f(−3) = 0
−27 + 9a + 21 + b = 0 → 9a + b = 6 ... (2)
−27 + 9a + 21 + b = 0 → 9a + b = 6 ... (2)
3
Subtract (1) from (2): 5a = −6 → a = −6/5...
Let me use cleaner values: try f(2)=6: 8+4a−14+b=6 → 4a+b=12, and f(−3)=0: −27+9a+21+b=0 → 9a+b=6.
Subtracting: 5a=−6 → a=−6/5. Not integer. Revised example: remainder when dividing by (x−2) is 4 (not 6):
4a+b=10 and 9a+b=6 → 5a=−4 → still not clean.
Revised: f(2)=0 (factor) and f(−3) gives remainder 4:
f(2)=0: 8+4a−14+b=0 → 4a+b=6 ...(1)
f(−3)=4: −27+9a+21+b=4 → 9a+b=10 ...(2)
(2)−(1): 5a=4 → still not clean. Use a=2, b=−2 directly:
Clean version: a=1, b=−6:
f(x) = x³+x²−7x−6 (well known factorisation: try f(−1)=−1+1+7−6=1≠0; f(2)=8+4−14−6=−8≠0; f(3)=27+9−21−6=9≠0; f(−2)=−8+4+14−6=4≠0. Try (x+1): f(−1)=−1+1+7−6=1≠0.)
Use standard example: f(x) = 2x³−3x²−11x+6, a=−3/2, b=3, x=−2: f(−2)=−16−12+22+6=0 ✓
So (x+2) is a factor. Divide: 2x³−3x²−11x+6 ÷ (x+2) = 2x²−7x+3 = (2x−1)(x−3)
f(x) = (x+2)(2x−1)(x−3)
Let me use cleaner values: try f(2)=6: 8+4a−14+b=6 → 4a+b=12, and f(−3)=0: −27+9a+21+b=0 → 9a+b=6.
Subtracting: 5a=−6 → a=−6/5. Not integer. Revised example: remainder when dividing by (x−2) is 4 (not 6):
4a+b=10 and 9a+b=6 → 5a=−4 → still not clean.
Revised: f(2)=0 (factor) and f(−3) gives remainder 4:
f(2)=0: 8+4a−14+b=0 → 4a+b=6 ...(1)
f(−3)=4: −27+9a+21+b=4 → 9a+b=10 ...(2)
(2)−(1): 5a=4 → still not clean. Use a=2, b=−2 directly:
Clean version: a=1, b=−6:
f(x) = x³+x²−7x−6 (well known factorisation: try f(−1)=−1+1+7−6=1≠0; f(2)=8+4−14−6=−8≠0; f(3)=27+9−21−6=9≠0; f(−2)=−8+4+14−6=4≠0. Try (x+1): f(−1)=−1+1+7−6=1≠0.)
Use standard example: f(x) = 2x³−3x²−11x+6, a=−3/2, b=3, x=−2: f(−2)=−16−12+22+6=0 ✓
So (x+2) is a factor. Divide: 2x³−3x²−11x+6 ÷ (x+2) = 2x²−7x+3 = (2x−1)(x−3)
f(x) = (x+2)(2x−1)(x−3)
📐 Worked Example 7 — Finding Unknown Coefficients
g(x) = 2x³ + px² + qx − 6. Given that (x−1) and (x+2) are both factors of g(x), find p and q. Hence fully factorise g(x).
1
g(1) = 0: 2 + p + q − 6 = 0 → p + q = 4 ... (1)
2
g(−2) = 0: −16 + 4p − 2q − 6 = 0 → 4p − 2q = 22 → 2p − q = 11 ... (2)
3
Add (1) and (2): 3p = 15 → p = 5. From (1): q = 4−5 = −1
4
g(x) = 2x³+5x²−x−6. Since (x−1)(x+2) are factors, divide by (x−1)(x+2) = x²+x−2:
2x³+5x²−x−6 ÷ (x²+x−2) = 2x+3
g(x) = (x−1)(x+2)(2x+3)
2x³+5x²−x−6 ÷ (x²+x−2) = 2x+3
g(x) = (x−1)(x+2)(2x+3)
3. Partial Fractions
Partial Fractions: Expressing a rational function f(x)/g(x) as a sum of simpler fractions. Essential for integration in Lesson 7 and for binomial expansions in Lesson 3. The degree of the numerator must be LESS than the degree of the denominator — if not, perform polynomial division first.
Three Types of Partial Fractions
| Type | Form of Denominator | Partial Fraction Form |
|---|---|---|
| Type 1 | Two distinct linear factors (ax+b)(cx+d) |
A/(ax+b) + B/(cx+d) |
| Type 2 | Repeated linear factor (ax+b)²(cx+d) |
A/(ax+b) + B/(ax+b)² + C/(cx+d) |
| Type 3 | Irreducible quadratic factor (ax²+bx+c)(dx+e) |
(Ax+B)/(ax²+bx+c) + C/(dx+e) |
📐 Worked Example 8 — Type 1: Two Distinct Linear Factors
Express (5x + 1)/[(x+1)(x−2)] as partial fractions.
1
Write: (5x+1)/[(x+1)(x−2)] = A/(x+1) + B/(x−2)
2
Multiply both sides by (x+1)(x−2):
5x+1 = A(x−2) + B(x+1)
5x+1 = A(x−2) + B(x+1)
3
Cover-up / substitution method:
Let x = 2: 11 = A(0) + B(3) → B = 11/3
Let x = −1: −4 = A(−3) + B(0) → A = 4/3
Let x = 2: 11 = A(0) + B(3) → B = 11/3
Let x = −1: −4 = A(−3) + B(0) → A = 4/3
4
(5x+1)/[(x+1)(x−2)] = 4/[3(x+1)] + 11/[3(x−2)]
📐 Worked Example 9 — Type 2: Repeated Linear Factor
Express (3x² + x + 1)/[(x−1)²(x+2)] as partial fractions.
1
Write: (3x²+x+1)/[(x−1)²(x+2)] = A/(x−1) + B/(x−1)² + C/(x+2)
2
Multiply through by (x−1)²(x+2):
3x²+x+1 = A(x−1)(x+2) + B(x+2) + C(x−1)²
3x²+x+1 = A(x−1)(x+2) + B(x+2) + C(x−1)²
3
Let x = 1: 5 = A(0) + B(3) + C(0) → B = 5/3
Let x = −2: 11 = A(0) + B(0) + C(9) → C = 11/9
Let x = −2: 11 = A(0) + B(0) + C(9) → C = 11/9
4
Compare coefficients of x² (or use x=0) to find A:
x=0: 1 = A(−1)(2) + B(2) + C(1) = −2A + 10/3 + 11/9
1 = −2A + 30/9 + 11/9 = −2A + 41/9
2A = 41/9−1 = 32/9 → A = 16/9
x=0: 1 = A(−1)(2) + B(2) + C(1) = −2A + 10/3 + 11/9
1 = −2A + 30/9 + 11/9 = −2A + 41/9
2A = 41/9−1 = 32/9 → A = 16/9
5
(3x²+x+1)/[(x−1)²(x+2)] = 16/[9(x−1)] + 5/[3(x−1)²] + 11/[9(x+2)]
📐 Worked Example 10 — Improper Fraction (Degree ≥ Degree)
Express (2x² + 3x − 1)/[(x+1)(x−2)] as partial fractions.
1
Degree of numerator (2) = degree of denominator (2) → improper. Divide first.
(x+1)(x−2) = x²−x−2. Divide 2x²+3x−1 by x²−x−2:
2x²+3x−1 = 2(x²−x−2) + (5x+3)
So: (2x²+3x−1)/[(x+1)(x−2)] = 2 + (5x+3)/[(x+1)(x−2)]
(x+1)(x−2) = x²−x−2. Divide 2x²+3x−1 by x²−x−2:
2x²+3x−1 = 2(x²−x−2) + (5x+3)
So: (2x²+3x−1)/[(x+1)(x−2)] = 2 + (5x+3)/[(x+1)(x−2)]
2
Now decompose (5x+3)/[(x+1)(x−2)] = A/(x+1) + B/(x−2):
5x+3 = A(x−2)+B(x+1)
x=2: 13 = 3B → B = 13/3
x=−1: −2 = −3A → A = 2/3
5x+3 = A(x−2)+B(x+1)
x=2: 13 = 3B → B = 13/3
x=−1: −2 = −3A → A = 2/3
3
(2x²+3x−1)/[(x+1)(x−2)] = 2 + 2/[3(x+1)] + 13/[3(x−2)]
Partial Fractions Strategy:
1. Check if improper (degree numerator ≥ degree denominator) → divide first.
2. Factorise the denominator completely.
3. Write the correct form of partial fractions for each factor type.
4. Multiply through and use substitution (cover-up method) for distinct linear factors.
5. For repeated or quadratic factors, also compare coefficients.
6. Always verify by recombining over the common denominator.
1. Check if improper (degree numerator ≥ degree denominator) → divide first.
2. Factorise the denominator completely.
3. Write the correct form of partial fractions for each factor type.
4. Multiply through and use substitution (cover-up method) for distinct linear factors.
5. For repeated or quadratic factors, also compare coefficients.
6. Always verify by recombining over the common denominator.
4. Simultaneous Equations — One Linear, One Quadratic
When one equation is linear and one is quadratic, substitute the linear equation into the quadratic. The discriminant of the resulting equation tells you how many intersection points exist.
Number of Intersections — Line and Curve:
After substitution, get ax²+bx+c = 0.
Δ > 0 → two distinct intersection points (line crosses curve twice)
Δ = 0 → one intersection (line is tangent to curve)
Δ < 0 → no intersection (line misses curve entirely)
After substitution, get ax²+bx+c = 0.
Δ > 0 → two distinct intersection points (line crosses curve twice)
Δ = 0 → one intersection (line is tangent to curve)
Δ < 0 → no intersection (line misses curve entirely)
📐 Worked Example 11 — Line Tangent to a Curve
Find the values of k for which the line y = kx + 3 is tangent to the curve y = x² + 2x − 1.
1
Substitute y = kx+3 into y = x²+2x−1:
kx+3 = x²+2x−1 → x²+(2−k)x−4 = 0
kx+3 = x²+2x−1 → x²+(2−k)x−4 = 0
2
For tangency, Δ = 0:
(2−k)²−4(1)(−4) = 0
4−4k+k²+16 = 0
k²−4k+20 = 0
(2−k)²−4(1)(−4) = 0
4−4k+k²+16 = 0
k²−4k+20 = 0
3
Δ of this: 16−80 = −64 < 0 → no real solutions.
This means there is no real value of k for which the line is tangent!
(Revised: try y=kx+3 tangent to y=x²−4x+1)
kx+3 = x²−4x+1 → x²−(4+k)x−2=0
Δ=0: (4+k)²+8=0 → (4+k)²=−8 → still no real k.
Working example: Line y=kx−5 tangent to y=x²:
kx−5=x² → x²−kx+5=0. Δ=k²−20=0 → k = ±2√5
This means there is no real value of k for which the line is tangent!
(Revised: try y=kx+3 tangent to y=x²−4x+1)
kx+3 = x²−4x+1 → x²−(4+k)x−2=0
Δ=0: (4+k)²+8=0 → (4+k)²=−8 → still no real k.
Working example: Line y=kx−5 tangent to y=x²:
kx−5=x² → x²−kx+5=0. Δ=k²−20=0 → k = ±2√5
5. Algebraic Identities and Proof
Identity (≡): An equation that is true for ALL values of the variable. Written with ≡ rather than =. To prove an identity, work on one side only (or transform both sides separately to a common form — never "cross-multiply" an identity).
📐 Worked Example 12 — Proving an Identity
Prove that (x+1)³ − (x−1)³ ≡ 2(3x²+1).
1
Expand LHS: (x+1)³ = x³+3x²+3x+1. (x−1)³ = x³−3x²+3x−1
2
LHS = (x³+3x²+3x+1) − (x³−3x²+3x−1)
= x³+3x²+3x+1−x³+3x²−3x+1
= 6x²+2 = 2(3x²+1) = RHS ✓
= x³+3x²+3x+1−x³+3x²−3x+1
= 6x²+2 = 2(3x²+1) = RHS ✓
📐 Worked Example 13 — Polynomial Identity (Comparing Coefficients)
Given that 2x³+7x²+2x−3 ≡ (2x−1)(x²+ax+b) + c for all x, find a, b, and c.
1
Expand RHS: (2x−1)(x²+ax+b)+c = 2x³+2ax²+2bx−x²−ax−b+c
= 2x³+(2a−1)x²+(2b−a)x+(−b+c)
= 2x³+(2a−1)x²+(2b−a)x+(−b+c)
2
Compare coefficients:
x²: 2a−1 = 7 → a = 4
x¹: 2b−a = 2 → 2b−4 = 2 → b = 3
x⁰: −b+c = −3 → −3+c = −3 → c = 0
x²: 2a−1 = 7 → a = 4
x¹: 2b−a = 2 → 2b−4 = 2 → b = 3
x⁰: −b+c = −3 → −3+c = −3 → c = 0
📝 Exam Practice Questions
Q1 [4 marks] — Find the values of k for which x² + (k−2)x + (2k+1) = 0 has real roots. For what value of k does the equation have equal roots?
For real roots: Δ ≥ 0
Δ = (k−2)² − 4(2k+1) = k²−4k+4−8k−4 = k²−12k
k²−12k ≥ 0 → k(k−12) ≥ 0
k ≤ 0 or k ≥ 12
Equal roots: Δ = 0 → k(k−12) = 0 → k = 0 or k = 12
Δ = (k−2)² − 4(2k+1) = k²−4k+4−8k−4 = k²−12k
k²−12k ≥ 0 → k(k−12) ≥ 0
k ≤ 0 or k ≥ 12
Equal roots: Δ = 0 → k(k−12) = 0 → k = 0 or k = 12
Exam Tip: "Real roots" means Δ ≥ 0 (includes equal roots). "Distinct real roots" means Δ > 0. "Equal/repeated roots" means Δ = 0. Know the difference — these are very commonly tested.
Q2 [3 marks] — Solve the inequality 2x² − 7x − 4 < 0.
Factorise: 2x²−7x−4 = (2x+1)(x−4)
Roots: x = −½ and x = 4. U-shaped parabola (a=2>0).
Parabola is below x-axis BETWEEN the roots:
−½ < x < 4
Roots: x = −½ and x = 4. U-shaped parabola (a=2>0).
Parabola is below x-axis BETWEEN the roots:
−½ < x < 4
Q3 [4 marks] — f(x) = x³ + 3x² + px + q. Given that (x+2) is a factor and that when f(x) is divided by (x−1) the remainder is 12, find p and q. Hence fully factorise f(x).
f(−2) = 0: −8+12−2p+q = 0 → −2p+q = −4 ... (1)
f(1) = 12: 1+3+p+q = 12 → p+q = 8 ... (2)
(2)−(1): 3p = 12 → p = 4. From (2): q = 4
f(x) = x³+3x²+4x+4. (x+2) is a factor → divide:
x³+3x²+4x+4 ÷ (x+2) = x²+x+2
Δ of x²+x+2: 1−8=−7<0 → no further real factors
f(x) = (x+2)(x²+x+2)
f(1) = 12: 1+3+p+q = 12 → p+q = 8 ... (2)
(2)−(1): 3p = 12 → p = 4. From (2): q = 4
f(x) = x³+3x²+4x+4. (x+2) is a factor → divide:
x³+3x²+4x+4 ÷ (x+2) = x²+x+2
Δ of x²+x+2: 1−8=−7<0 → no further real factors
f(x) = (x+2)(x²+x+2)
Q4 [4 marks] — Express (4x² − 3x + 5)/[(2x+1)(x−2)²] as partial fractions.
Write: A/(2x+1) + B/(x−2) + C/(x−2)²
4x²−3x+5 = A(x−2)² + B(2x+1)(x−2) + C(2x+1)
x=2: 16−6+5=15 = C(5) → C = 3
x=−½: 1+3/2+5=15/2 = A(25/4) → A = (15/2)×(4/25) = A = 6/5
Compare x²: 4 = A+2B → 4 = 6/5+2B → 2B = 14/5 → B = 7/5
Result: 6/[5(2x+1)] + 7/[5(x−2)] + 3/(x−2)²
4x²−3x+5 = A(x−2)² + B(2x+1)(x−2) + C(2x+1)
x=2: 16−6+5=15 = C(5) → C = 3
x=−½: 1+3/2+5=15/2 = A(25/4) → A = (15/2)×(4/25) = A = 6/5
Compare x²: 4 = A+2B → 4 = 6/5+2B → 2B = 14/5 → B = 7/5
Result: 6/[5(2x+1)] + 7/[5(x−2)] + 3/(x−2)²
Q5 [3 marks] — Show that x² + 4x + 7 > 0 for all real values of x.
Method 1 — Discriminant:
a = 1 > 0 (opens upward). Δ = 16−28 = −12 < 0.
Since a > 0 and Δ < 0, the parabola opens upward and never crosses the x-axis.
Therefore x²+4x+7 > 0 for all x ∈ ℝ. ✓
Method 2 — Completing the square:
x²+4x+7 = (x+2)²+3. Since (x+2)² ≥ 0 for all x, (x+2)²+3 ≥ 3 > 0. ✓
a = 1 > 0 (opens upward). Δ = 16−28 = −12 < 0.
Since a > 0 and Δ < 0, the parabola opens upward and never crosses the x-axis.
Therefore x²+4x+7 > 0 for all x ∈ ℝ. ✓
Method 2 — Completing the square:
x²+4x+7 = (x+2)²+3. Since (x+2)² ≥ 0 for all x, (x+2)²+3 ≥ 3 > 0. ✓
Exam Tip: "Show that ... > 0 for all real x" — use EITHER the discriminant method or completing the square. Both are equally valid. Completing the square gives the stronger result (shows the minimum value).
Q6 [3 marks] — Given that 3x³−2x²+5x−4 ≡ (x−1)(ax²+bx+c)+d for all x, find a, b, c, and d.
Expand: (x−1)(ax²+bx+c)+d = ax³+bx²+cx−ax²−bx−c+d
= ax³+(b−a)x²+(c−b)x+(d−c)
Compare coefficients:
x³: a = 3
x²: b−a = −2 → b−3 = −2 → b = 1
x¹: c−b = 5 → c−1 = 5 → c = 6
x⁰: d−c = −4 → d−6 = −4 → d = 2
Note: d=2 confirms remainder when dividing by (x−1) is 2. Check: f(1)=3−2+5−4=2 ✓
= ax³+(b−a)x²+(c−b)x+(d−c)
Compare coefficients:
x³: a = 3
x²: b−a = −2 → b−3 = −2 → b = 1
x¹: c−b = 5 → c−1 = 5 → c = 6
x⁰: d−c = −4 → d−6 = −4 → d = 2
Note: d=2 confirms remainder when dividing by (x−1) is 2. Check: f(1)=3−2+5−4=2 ✓