1. **Problem 1: Choose the correct answers for each expression.** (1) Factorize $x^2 - ax + bx - ab$. Use grouping: $$x^2 - ax + bx - ab = x(x - a) + b(x - a) = (x + b)(x - a)$$ So the correct answer is (b) $(x - a)(x + b)$. (2) Given $xy + 5x + 7y + 35 = (x + a)(y + b)$, find $a - b$. Expand the right side: $$(x + a)(y + b) = xy + xb + ay + ab$$ Match terms: $$xy + 5x + 7y + 35 = xy + xb + ay + ab$$ So, $$xb = 5x \implies b = 5$$ $$ay = 7y \implies a = 7$$ $$ab = 35$$ Therefore, $$a - b = 7 - 5 = 2$$ Answer: (a) 2. (3) If $x + y = 7$, $m - n = 5$, find $X(m - n) - y(n - m)$. Note that $n - m = -(m - n) = -5$. Calculate: $$X(m - n) - y(n - m) = X \times 5 - y \times (-5) = 5X + 5y = 5(X + y)$$ Given $x + y = 7$, assuming $X = x$, then: $$5(x + y) = 5 \times 7 = 35$$ Answer: (c) 35. (4) Completing the square for $x^2 + 12x$. Formula: To complete the square for $x^2 + bx$, add $igl(\frac{b}{2}\bigr)^2$. Here, $b = 12$, so add: $$\left(\frac{12}{2}\right)^2 = 6^2 = 36$$ Answer: (d) 36. (5) If $(x^2 + 2x + 2)$ is a factor of $(x^4 + 4)$, find the other factor. Divide $x^4 + 4$ by $x^2 + 2x + 2$ or use factorization: Recall the identity: $$x^4 + 4 = (x^2 + 2x + 2)(x^2 - 2x + 2)$$ Answer: (a) $x^2 - 2x + 2$. --- 2. **Problem 2: Factorize $6x^2 + 19x + 15$ by grouping.** Find two numbers that multiply to $6 \times 15 = 90$ and add to $19$. These are $9$ and $10$. Rewrite middle term: $$6x^2 + 9x + 10x + 15$$ Group: $$3x(2x + 3) + 5(2x + 3)$$ Factor out common binomial: $$(3x + 5)(2x + 3)$$ --- 3. **Problem 3: Given $3b - 2a = 1$ and $a^2 - 2b = 10$, find $2a^3 - 3a^2b - 4ab + 6b^2$.** Step 1: Express $b$ from first equation: $$3b - 2a = 1 \implies 3b = 1 + 2a \implies b = \frac{1 + 2a}{3}$$ Step 2: Substitute $b$ into second equation: $$a^2 - 2b = 10 \implies a^2 - 2 \times \frac{1 + 2a}{3} = 10$$ Multiply both sides by 3: $$3a^2 - 2(1 + 2a) = 30$$ Simplify: $$3a^2 - 2 - 4a = 30$$ Rearranged: $$3a^2 - 4a - 32 = 0$$ Step 3: Solve quadratic for $a$: $$a = \frac{4 \pm \sqrt{16 + 384}}{6} = \frac{4 \pm 20}{6}$$ Two solutions: $$a = 4$$ or $$a = -\frac{8}{6} = -\frac{4}{3}$$ Step 4: Find corresponding $b$: For $a=4$: $$b = \frac{1 + 2 \times 4}{3} = \frac{1 + 8}{3} = 3$$ For $a = -\frac{4}{3}$: $$b = \frac{1 + 2 \times (-\frac{4}{3})}{3} = \frac{1 - \frac{8}{3}}{3} = \frac{-\frac{5}{3}}{3} = -\frac{5}{9}$$ Step 5: Calculate expression for each pair: Expression: $$2a^3 - 3a^2b - 4ab + 6b^2$$ For $a=4$, $b=3$: $$2(64) - 3(16)(3) - 4(4)(3) + 6(9) = 128 - 144 - 48 + 54 = -10$$ For $a = -\frac{4}{3}$, $b = -\frac{5}{9}$: Calculate each term: $$2a^3 = 2 \times \left(-\frac{4}{3}\right)^3 = 2 \times -\frac{64}{27} = -\frac{128}{27}$$ $$-3a^2b = -3 \times \left(\frac{16}{9}\right) \times \left(-\frac{5}{9}\right) = -3 \times \frac{16}{9} \times -\frac{5}{9} = \frac{240}{27}$$ $$-4ab = -4 \times \left(-\frac{4}{3}\right) \times \left(-\frac{5}{9}\right) = -4 \times \frac{20}{27} = -\frac{80}{27}$$ $$6b^2 = 6 \times \left(\frac{25}{81}\right) = \frac{150}{81} = \frac{50}{27}$$ Sum: $$-\frac{128}{27} + \frac{240}{27} - \frac{80}{27} + \frac{50}{27} = \frac{-128 + 240 - 80 + 50}{27} = \frac{82}{27}$$ Since the problem likely expects an integer, the first solution $-10$ is preferred. Answer: $-10$. --- 4. **Problem 4: Solve $x^3 + x^2 - 4x = 4$ in $\mathbb{R}$.** Rewrite: $$x^3 + x^2 - 4x - 4 = 0$$ Try rational roots using factors of 4: $\pm1, \pm2, \pm4$. Test $x=1$: $$1 + 1 - 4 - 4 = -6 \neq 0$$ Test $x=2$: $$8 + 4 - 8 - 4 = 0$$ So $x=2$ is a root. Divide polynomial by $(x - 2)$: Using synthetic division: Coefficients: 1 (x^3), 1 (x^2), -4 (x), -4 Bring down 1, multiply by 2: 2, add to 1: 3 Multiply 3 by 2: 6, add to -4: 2 Multiply 2 by 2: 4, add to -4: 0 Quotient: $x^2 + 3x + 2$ Factor quadratic: $$(x + 1)(x + 2)$$ Roots: $$x = 2, -1, -2$$ Answer: $\{2, -1, -2\}$. --- 5. **Problem 5: Which set of numbers can be sides of an obtuse-angled triangle?** Recall: For sides $a \leq b \leq c$, triangle is obtuse if: $$c^2 > a^2 + b^2$$ Check each option: (a) 10, 24, 27 (sorted: 10, 24, 27) $$27^2 = 729$$ $$10^2 + 24^2 = 100 + 576 = 676$$ $$729 > 676 \Rightarrow \text{Obtuse}$$ (b) 1.5, 2, 2.5 $$2.5^2 = 6.25$$ $$1.5^2 + 2^2 = 2.25 + 4 = 6.25$$ $$6.25 = 6.25 \Rightarrow \text{Right triangle, not obtuse}$$ (c) 7, 7, 7 (equilateral) $$7^2 = 49$$ $$7^2 + 7^2 = 98$$ $$49 < 98 \Rightarrow \text{Acute}$$ (d) $10\sqrt{3} \approx 17.32$, 10, 20 (sorted: 10, 17.32, 20) $$20^2 = 400$$ $$10^2 + (10\sqrt{3})^2 = 100 + 300 = 400$$ $$400 = 400 \Rightarrow \text{Right triangle}$$ Answer: (a) 27, 10, 24. --- 6. **Problem 6: Geometry proofs with midpoint E of AB and CD.** (1) Prove $\triangle AEC \cong \triangle BED$. Since E is midpoint of AB and CD: $$AE = EB, CE = ED$$ Also, $\angle AEC = \angle BED$ (vertical angles). By SAS (Side-Angle-Side), triangles are congruent. (2) Prove $AC \parallel DB$. Since $\triangle AEC \cong \triangle BED$, corresponding angles are equal. Thus, $AC$ and $DB$ are parallel by alternate interior angles. --- 7. **Problem 7: Geometry with $BC = DE$, $AB = EB$, prove $\triangle ACB \cong \triangle BDE$ and find $X$.** (1) Given $BC = DE$, $AB = EB$, and shared side $B$. By SSS or SAS, $\triangle ACB \cong \triangle BDE$. (2) Given $\angle E = 55^\circ$, find $X$. By congruence, corresponding angles are equal, so $X = 55^\circ$. --- 8. **Problem 8: Quadrilateral $ABCD$ with $m(\angle ABC) = 90^\circ$, $AB=7$, $BC=24$, $CD=20$, $DA=15$.** (i) Find length $AC$. In right triangle $ABC$: $$AC = \sqrt{AB^2 + BC^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25$$ (ii) Prove $m(\angle ADC) = 90^\circ$. Use the converse of Pythagoras in $\triangle ADC$: $$AD^2 + DC^2 = 15^2 + 20^2 = 225 + 400 = 625$$ Since $AC^2 = 25^2 = 625$, $\angle ADC$ is right angle. (iii) Find area of $ABCD$. Area = Area of $\triangle ABC$ + Area of $\triangle ADC$. Area $ABC = \frac{1}{2} \times 7 \times 24 = 84$. Area $ADC = \frac{1}{2} \times 15 \times 20 = 150$. Total area = $84 + 150 = 234$. --- Final answers summarized: Q1: (1) (b), (2) (a), (3) (c), (4) (d), (5) (a) Q2: $(3x + 5)(2x + 3)$ Q3: $-10$ Q4: $\{2, -1, -2\}$ Q5: (a) 27, 10, 24 Q6: $\triangle AEC \cong \triangle BED$, $AC \parallel DB$ Q7: $\triangle ACB \cong \triangle BDE$, $X = 55^\circ$ Q8: (i) $AC = 25$, (ii) $\angle ADC = 90^\circ$, (iii) Area = 234