Polynomials

11

Step 1: Compare x³+3x²-x-3 with ax³+bx²+cx+d: a=1, b=3, c=-1, d=-3. Step 2: Sum of zeroes: α+β+γ = -b/a = -3. Step 3: Sum of products taken two at a time: αβ+βγ+γα = c/a = -1. Step 4: Use identity: α²+β²+γ² = (α+β+γ)² - 2(αβ+βγ+γα) = (-3)² - 2(-1) = 9 + 2 = 11. Common mistake: Students often use (α+β+γ)² = α²+β²+γ² forgetting the cross-product terms. The identity (α+β+γ)² = α²+β²+γ² + 2(αβ+βγ+γα) is essential.

x² - x + 1

Step 1: Use the Division Algorithm: Dividend = Divisor × Quotient + Remainder. Step 2: Dividend - Remainder = Divisor × Quotient. So (x³-3x²+x+2) - (-2x+4) = g(x)(x-2). Step 3: x³-3x²+x+2+2x-4 = x³-3x²+3x-2. Step 4: g(x) = (x³-3x²+3x-2)/(x-2). Dividing: x³-3x²+3x-2 ÷ (x-2) = x²-x+1 [since (x-2)(x²-x+1) = x³-x²+x-2x²+2x-2 = x³-3x²+3x-2]. Step 5: g(x) = x²-x+1. Common mistake: Not subtracting the remainder before dividing; directly dividing dividend by quotient.

-1

Step 1: For the cubic polynomial ax³+3x²-13x+6, the sum of zeroes = -b/a = -3/a. Step 2: We are given that sum of zeroes = 3. Step 3: Set up equation: -3/a = 3. Step 4: Solving: a = -3/3 = -1. Step 5: Verify — with a=-1: polynomial is -x³+3x²-13x+6. Sum of zeroes = -3/(-1) = 3 ✓. Common mistake: Using sum formula as +b/a instead of -b/a. Always remember sum of zeroes of ax³+bx²+cx+d is -b/a.

Other zero = -2, λ = -1

Step 1: Let α = 1/2 and β be the other zero. From 2x²+3x+λ: sum of zeroes = -3/2, product = λ/2. Step 2: Find β using sum: 1/2 + β = -3/2, so β = -3/2 - 1/2 = -4/2 = -2. Step 3: Find λ using product: (1/2)(-2) = λ/2, so -1 = λ/2, giving λ = -2. Wait — rechecking: product = λ/2, (1/2)(-2) = -1 = λ/2, so λ = -2. Common mistake causes: verifying p(1/2) = 2(1/4)+3(1/2)+λ = 1/2+3/2+λ = 2+λ = 0, so λ = -2. Other zero = -2, λ = -2. Correct answer should be λ = -2. Note discrepancy in options — the correct mathematical answer is λ = -2 and other zero = -2.