Polynomials

(x-3)(x+2)(x-1)(x+1)

Step 1: Since (x-3) is a factor, divide p(x) by (x-3) using synthetic or long division. Step 2: x⁴ - x³ - 7x² + x + 6 ÷ (x-3) = x³ + 2x² - x - 2. Step 3: Now factorise x³ + 2x² - x - 2. Try x=1: 1+2-1-2=0 ✓. So (x-1) is a factor. Step 4: x³ + 2x² - x - 2 ÷ (x-1) = x² + 3x + 2 = (x+1)(x+2). Step 5: Therefore p(x) = (x-3)(x-1)(x+1)(x+2). Common mistake: students often stop after finding one factor and do not fully factorise. Always continue until you reach linear factors.

All real values of k

Step 1: By Factor Theorem, (x+2) is a factor if and only if p(-2) = 0. Step 2: p(-2) = k(-2)³ + 5(-2)² - 2k(-2) - 10 Step 3: = -8k + 20 + 4k - 10 = -4k + 10. Step 4: Wait: -8k + 4k = -4k, and 20 - 10 = 10. So p(-2) = -4k + 10. Step 5: Setting -4k + 10 = 0 gives k = 10/4 = 5/2, not 'all values'. Let me re-examine the polynomial: kx³ + 5x² - 2kx - 10. Group: x²(kx+5) - 2(kx+5) = (kx+5)(x²-2). This factors for all k, but (x+2) divides it only if p(-2)=0: k=5/2. The correct answer is k=5/2. Among the options, none exactly states 5/2; however, the closest instructional point is that students must n

12

Step 1: We know the identity: (x+y+z)² = x²+y²+z² + 2(xy+yz+zx). Step 2: Substitute the given values: (x+y+z)² = 50 + 2(47). Step 3: (x+y+z)² = 50 + 94 = 144. Step 4: Taking the positive square root: x+y+z = √144 = 12. Step 5: The answer is 12. Common mistake: students add xy+yz+zx directly without multiplying by 2. Remember the identity includes 2(xy+yz+zx), not just (xy+yz+zx).

110

Step 1: We are given x + 1/x = 5. We need x³ + 1/x³. Step 2: First find x² + 1/x²: (x + 1/x)² = x² + 2 + 1/x² = 25, so x² + 1/x² = 23. Step 3: Use the identity: x³ + 1/x³ = (x + 1/x)(x² - 1 + 1/x²). Step 4: = (x + 1/x)[(x² + 1/x²) - 1] = 5 × (23 - 1) = 5 × 22 = 110. Step 5: The answer is 110. Students often mistakenly cube (x+1/x) directly and get 125, forgetting the identity structure. The key step is to first find x²+1/x² before computing the cube.