Expressions and proof, complex numbers and equations (Mathematics II)

Substituting a=x and b=2 into (a+b)³=a³+3a²b+3ab²+b³ gives x³+6x²+12x+8. The option x³+8 is the classic slip of forgetting the two middle terms 3a²b and 3ab². Check the run of coefficients 1,3,3,1.

By a³−b³=(a−b)(a²+ab+b²) we get x³−27=(x−3)(x²+3x+9). The middle term of the second bracket is +ab, with the sign opposite to that in the first bracket. The option (x−3)(x²−3x+9) reverses that sign, and (x−3)³ expands to x³−9x²+27x−27, which does not match.

The general term is C(5,r)·x⁵⁻ʳ·2ʳ. The term in x³ occurs at r=2, and its coefficient is C(5,2)·2² = 10×4 = 40. The value 80 multiplies by 2³, 20 multiplies by 2¹, and 10 forgets to multiply by 2² at all.

The general term is C(6,r)·(2x)⁶⁻ʳ·(−y)ʳ. The term in x⁴y² occurs at r=2, giving C(6,2)·2⁴·(−1)² = 15×16×1 = 240. The value −240 mishandles the sign of (−y)², 60 uses 2² instead of 2⁴, and 480 multiplies by 2⁵.

The numerator is (x+3)(x−3) and the denominator is (x+3)(x−2). Cancelling the common factor x+3 gives (x−3)/(x−2). The golden rule is to factorise numerator and denominator first. Since x−3 and x−2 are different factors, nothing more cancels.

Putting them over a common denominator gives {2(x+1)+3(x−1)}/{(x−1)(x+1)} = (2x+2+3x−3)/(x²−1) = (5x−1)/(x²−1). The option 5/(x²−1) simply adds 2+3 without adjusting the numerators, and (5x+1) is a sign error.

The left side is (a+b)x+(a−b). Comparing coefficients gives a+b=3 and a−b=5. Adding them gives 2a=8, so a=4 and b=3−4=−1. The option a=4, b=1 swaps a−b=5 with a+b=3. Substituting values also works: x=1 gives 2a=8 and x=−1 gives −2b=2.

Substituting x=1 gives c=1. Comparing coefficients of x² gives a=1. Substituting x=0 gives 0=a−b+c=1−b+1, so b=2. Checking: (x−1)²+2(x−1)+1 = x²−2x+1+2x−2+1 = x², which holds. A sign error on b, giving b=−2, is the typical mistake.

Since x>0 and 9/x>0, the arithmetic-geometric mean inequality gives x+9/x ≧ 2√(x·9/x) = 2√9 = 6. Equality holds when x=9/x, that is x=3. The option 3 takes only √9 and forgets to multiply by 2, and 9 confuses this with the product x·(9/x).

The arithmetic-geometric mean inequality gives x+y ≧ 2√(xy) = 2√4 = 4. Equality holds when x=y=2, which does satisfy xy=4, so the minimum is 4. The option 2 takes only √4 and forgets to multiply by 2, and 8 comes from computing 2×4.

a³+b³ = (a+b)³−3ab(a+b) = 27−3·1·3 = 27−9 = 18. The option 27 forgets the term −3ab(a+b), 24 subtracts only ab as 27−3, and 30 comes from wrongly taking a²+b² to be 9+2=11 and computing (a+b)(a²−ab+b²)=3×10.

The general term is C(6,r)·x⁶⁻ʳ·(1/x)ʳ = C(6,r)·x⁶⁻²ʳ. The constant term occurs when 6−2r=0, that is r=3, giving C(6,3)=20. The value 15 is C(6,2) and 6 is C(6,1). The method is to find first the r that makes the exponent of x zero.

Add real parts to real parts and imaginary parts to imaginary parts: (3+1)+(2−4)i = 4−2i. The option 4+2i mishandles the sign of the imaginary part, 2−2i miscomputes the real part, and 4−6i takes 2−4 to be −6.

Expanding gives 6−4i+3i−2i². Since i²=−1, the term −2i² becomes +2, so we get 6−i+2 = 8−i. The option 4−i takes i²=+1 by mistake, and 6−i drops the −2i² term. Handling i² correctly is what decides the answer.

The powers of i repeat with period 4 as i, −1, −i, 1. Since 23 = 4×5+3, we get i²³ = i³ = i²·i = −i. The method is to judge by the remainder, here 3, when 23 is divided by 4.

Multiply numerator and denominator by the conjugate 3+i. The numerator is (1+i)(3+i) = 3+i+3i+i² = 2+4i and the denominator is 9−i² = 10, so we get (2+4i)/10 = (1+2i)/5. The option (2+i)/5 multiplies the numerator by 3−i instead, and taking the denominator as 9−1=8 gives (1+2i)/4.

Convert to i first: √(−8)=2√2i and √(−2)=√2i, so the product is 2√2·√2·i² = 4×(−1) = −4. Writing √(−8)×√(−2)=√16=4 is the classic slip of overlooking that √a·√b=√(ab) does not hold for negative numbers.

The quadratic formula gives x = (−2±√(4−20))/2 = (−2±√(−16))/2 = (−2±4i)/2 = −1±2i. The option −2±4i forgets to divide by 2, and 1±2i is a sign error. Using D/4 = 1−5 = −4 is quicker: x = −1±√(−4) = −1±2i.

The condition is D=0. From D = k²−4(k+3) = k²−4k−12 = (k−6)(k+2) = 0 we get k=6, −2. Expanding −4(k+3) as something other than −4k−12, with a sign error, leads to wrong answers such as k=−6, 2.

The discriminant is D = (−5)²−4·3·4 = 25−48 = −23 < 0, so there are two distinct imaginary roots. With real coefficients, imaginary roots always come in conjugate pairs, so one real root together with one imaginary root is impossible.

By the relations between roots and coefficients, α+β = −b/a = −(−5)/1 = 5 and αβ = c/a = 3. The option α+β=−5 is the classic slip of forgetting the minus sign in −b/a.

α²+β² = (α+β)²−2αβ = 25−2×3 = 19. The option 25 forgets −2αβ, 31 adds as 25+6, and 22 subtracts αβ only once as 25−3.

From the relations between roots and coefficients, α+β=3 and αβ=5. Then α³+β³ = (α+β)³−3αβ(α+β) = 27−3·5·3 = 27−45 = −18. The option 27 forgets −3αβ(α+β), 72 adds as 27+45, and 12 comes from 27−15, forgetting to multiply by (α+β).

Because the coefficients are real, the conjugate 3−2i is also a root. The sum of the roots is 6 = −p, so p=−6, and the product is 9−(2i)² = 9+4 = 13 = q. The option q=5 takes the product as 9−4, and p=6 is a sign error. The equation is x²−6x+13=0.

By the remainder theorem the remainder is P(2) = 8−8+6−1 = 5. The option −23 comes from computing P(−2), a sign misreading. To divide by x−2 you substitute x=2.

Since x+1 = x−(−1), the remainder is P(−1) = −2+1+5+3 = 7. The option 1 comes from computing P(1)=2+1−5+3. The most important point is the sign switch: dividing by x+1 means substituting x=−1.

By the factor theorem, x−k is a factor when P(k)=0. Here P(2) = 8−12−8+12 = 0, so x−2 is a factor. We have P(−3)=−30, P(4)=12 and P(−1)=12, none of which is 0. In fact P(x)=(x−2)(x+2)(x−3).

In synthetic division, multiplying by 2 and adding down the coefficients 1, −4, 1, 6 gives 1, −2, −3, 0. So the quotient is x²−2x−3 with remainder 0. The option x²−6x+13 is the quotient on division by x+2, using −2, and arises from misreading the sign.

Since P(1)=1−7+6=0, x−1 is a factor. Synthetic division gives the quotient x²+x−6=(x+3)(x−2). Hence (x−1)(x−2)(x+3)=0 and x=1, 2, −3. Checking: the product of the roots 1·2·(−3)=−6 agrees with minus the constant term.

Factorise x³−8=0 by the a³−b³ formula as (x−2)(x²+2x+4)=0. The roots of x²+2x+4=0 are x=−1±√3i, so the three solutions are x=2 and −1±√3i. Answering x=2 only drops the imaginary roots, and 1±√3i is a sign error in the quadratic formula.

By the factor theorem, P(1)=1−2+a+b=0 gives a+b=1, and P(−2)=−8−8−2a+b=0 gives −2a+b=16. Subtracting gives 3a=−15, so a=−5 and b=6. Writing +2a instead of −2a in P(−2) leads to a=15, b=−14.

x⁴−1 = (x²−1)(x²+1) = (x−1)(x+1)(x²+1). Since x²−1 can be split further, stopping at (x²−1)(x²+1) is incomplete. The factor x²+1 cannot be split over the real numbers; the form (x−i)(x+i) is a factorisation over the complex numbers.

Since ω is an imaginary root of x³−1=(x−1)(x²+x+1)=0, it satisfies x²+x+1=0. Hence ω²+ω+1=0 gives ω²+ω=−1. The two relations ω³=1 and ω²+ω+1=0 are the basic tools for computing with ω.

The remainder on division by a quadratic has degree at most 1, so write it as ax+b. Applying the remainder theorem to P(x)=(x−1)(x−2)Q(x)+ax+b gives P(1)=a+b=3 and P(2)=2a+b=5. Subtracting gives a=2 and b=1, so the remainder is 2x+1. Mixing up the simultaneous equations leads to −2x+7.

Because the coefficients are real, the conjugate 1−i is also a root. Writing the remaining root as γ, the sum of the three roots is (1+i)+(1−i)+γ = 2+γ = 4, so γ=2. Expanding (x²−2x+2)(x−2) = x³−4x²+6x−4 gives a=6 and b=−4. It is quick to use the fact that the sum of the roots equals the coefficient of x² with its sign reversed.