Chapter 2: Polynomials

Solution

−b/a = −(−5)/2 = 5/2

Solution

Zeroes = x-intercepts. No intersection → 0 real zeroes.

Solution

S = −3+2 = −1, P = −3×2 = −6. Polynomial: x²−(−1)x+(−6) = x²+x−6.

Check: −b/a = −1 = S ✓, c/a = −6 = P ✓.

Solution

Vertex below x-axis + opens upward → crosses x-axis at 2 points → 2 zeroes.

Solution

p(2) = 4+6+k = 0 → k = −10.

Solution

Passes through origin means p(0) = 0, so 0 is a zero.

Solution

α+β = 7/6, αβ = −1/2. α²+β² = (α+β)²−2αβ = 49/36+1 = 85/36.

Solution

α+β = 6. 3α+2β = 20 → 3α+2(6−α) = 20 → α = 8, β = −2. a = αβ = −16.

Solution

3x³÷x² = 3x → subtract 3x³+6x²+3x → −5x²−x+5.

−5x²÷x² = −5 → subtract −5x²−10x−5 → 9x+10.

Verify: (x²+2x+1)(3x−5)+9x+10 = 3x³+x²+2x+5 ✓

Solution

x²=3 → x=±√3. Sum = 0 = −b/a ✓. Product = −3 = c/a ✓.

Solution

α+β = k+6, αβ = 4k−2. Given k+6 = (4k−2)/2 = 2k−1 → k = 7.

Solution

6x²−9x+2x−3 = 3x(2x−3)+1(2x−3) = (3x+1)(2x−3).

Zeroes: −1/3, 3/2. Sum = 7/6 = −(−7)/6 ✓. Product = −1/2 = −3/6 ✓.

Solution

Dividing gives quotient 2t²+3t+4 with remainder 0. Hence t²−3 is a factor.

Solution

Zeroes equal and opposite → sum = 0. Sum = 2/k = 0 has no finite solution. Actually if zeroes are α and −α, sum = 0, so 2/k = 0 is impossible for finite k. The condition requires b = 0, but b = −2 ≠ 0. Alternatively, if the polynomial is rewritten with a leading coefficient, we need −(−2)/k = 0, impossible. However, examining the options: if k = 0, the polynomial becomes −2x which has zero at 0 (trivially equal and opposite to itself). k = 0 makes the polynomial linear with single zero 0.

Solution

g(x)(x−2) = x³−3x²+x+2−(−2x+4) = x³−3x²+3x−2.

Divide by (x−2): quotient x²−x+1, remainder 0. So g(x) = x²−x+1.

Solution

Factor from given zeroes: (x−2)²−3 = x²−4x+1.

Divide: x⁴−6x³−26x²+138x−35 ÷ (x²−4x+1) = x²−2x−35 = (x−7)(x+5).

Other zeroes: 7, −5.

Solution

α+β=5, αβ=6. New sum = (α+β)/(αβ) = 5/6. New product = 1/(αβ) = 1/6.

Polynomial: x²−(5/6)x+1/6 → 6x²−5x+1.

Solution

α+β = −5/2, αβ = k/2. α²+β²+αβ = (α+β)²−αβ = 25/4−k/2 = 21/4. So k/2 = 1, k = 2.

Solution

Sum = 3a = 3 → a = 1. Product = a(a²−b²) = −1 → 1−b² = −1 → b² = 2 → b = ±√2.

Solution

Division gives quotient 2x²+5, remainder x+2. So a=1, b=2, a+b=3.