0000:00:00:00:00Start34Number 34ExplanationA third-degree polynomial P(x)=x3+2x2+mx+nP(x) = x^3 + 2x^2 + mx + nP(x)=x3+2x2+mx+n divided by x2−4x+3x^2 - 4x + 3x2−4x+3 has a remainder of 3x+23x + 23x+2. Then the value of nnn = ....−20-20−20−16-16−16101010161616202020ExplanationThe basic concept of polynomial division is P(x)=g(x)⋅h(x)+s(x)P(x) = g(x) \cdot h(x) + s(x)P(x)=g(x)⋅h(x)+s(x) where P(x)P(x)P(x) is the polynomial being divided, g(x)g(x)g(x) is the divisor, h(x)h(x)h(x) is the quotient, and s(x)s(x)s(x) is the remainder. Thus x3+2x2+mx+n=(x2−4x+3)h(x)+(3x+2)x^3 + 2x^2 + mx + n = (x^2 - 4x + 3)h(x) + (3x + 2)x3+2x2+mx+n=(x2−4x+3)h(x)+(3x+2)x3+2x2+mx+n=(x−1)(x−3)h(x)+(3x+2)x^3 + 2x^2 + mx + n = (x - 1)(x - 3)h(x) + (3x + 2)x3+2x2+mx+n=(x−1)(x−3)h(x)+(3x+2) where P(x)=x3+2x2+mx+nP(x) = x^3 + 2x^2 + mx + nP(x)=x3+2x2+mx+n, g(x)=(x−1)(x−3)g(x) = (x - 1)(x - 3)g(x)=(x−1)(x−3), and s(x)=3x+2s(x) = 3x + 2s(x)=3x+2. Substitute x=1x = 1x=1 P(1)=(1−1)(1−3)h(1)+(3(1)+2)P(1) = (1 - 1)(1 - 3)h(1) + (3(1) + 2)P(1)=(1−1)(1−3)h(1)+(3(1)+2)13+2(1)2+m(1)+n=0+51^3 + 2(1)^2 + m(1) + n = 0 + 513+2(1)2+m(1)+n=0+51+2+m+n=51 + 2 + m + n = 51+2+m+n=5m+n=2…(1)m + n = 2 \quad \ldots (1)m+n=2…(1) Substitute x=3x = 3x=3 P(3)=(3−1)(3−3)h(3)+(3(3)+2)P(3) = (3 - 1)(3 - 3)h(3) + (3(3) + 2)P(3)=(3−1)(3−3)h(3)+(3(3)+2)33+2(3)2+m(3)+n=0+113^3 + 2(3)^2 + m(3) + n = 0 + 1133+2(3)2+m(3)+n=0+1127+18+3m+n=1127 + 18 + 3m + n = 1127+18+3m+n=1145+3m+n=1145 + 3m + n = 1145+3m+n=113m+n=−34…(2)3m + n = -34 \quad \ldots (2)3m+n=−34…(2) Eliminate both equations 3m+n=−34…(2)m+n=2…(1)(2)−(1):(3m+n)−(m+n)=−34−2−2m=36m=−18\begin{aligned} 3m + n &= -34 \quad \ldots (2) \\ m + n &= 2 \quad \ldots (1) \\ \hline (2) - (1): \quad (3m + n) - (m + n) &= -34 - 2 \\ -2m &= 36 \\ m &= -18 \end{aligned}3m+nm+n(2)−(1):(3m+n)−(m+n)−2mm=−34…(2)=2…(1)=−34−2=36=−18 Substitute into equation (1) m+n=2m + n = 2m+n=2−18+n=2-18 + n = 2−18+n=2n=20n = 20n=20
34Number 34ExplanationA third-degree polynomial P(x)=x3+2x2+mx+nP(x) = x^3 + 2x^2 + mx + nP(x)=x3+2x2+mx+n divided by x2−4x+3x^2 - 4x + 3x2−4x+3 has a remainder of 3x+23x + 23x+2. Then the value of nnn = ....−20-20−20−16-16−16101010161616202020ExplanationThe basic concept of polynomial division is P(x)=g(x)⋅h(x)+s(x)P(x) = g(x) \cdot h(x) + s(x)P(x)=g(x)⋅h(x)+s(x) where P(x)P(x)P(x) is the polynomial being divided, g(x)g(x)g(x) is the divisor, h(x)h(x)h(x) is the quotient, and s(x)s(x)s(x) is the remainder. Thus x3+2x2+mx+n=(x2−4x+3)h(x)+(3x+2)x^3 + 2x^2 + mx + n = (x^2 - 4x + 3)h(x) + (3x + 2)x3+2x2+mx+n=(x2−4x+3)h(x)+(3x+2)x3+2x2+mx+n=(x−1)(x−3)h(x)+(3x+2)x^3 + 2x^2 + mx + n = (x - 1)(x - 3)h(x) + (3x + 2)x3+2x2+mx+n=(x−1)(x−3)h(x)+(3x+2) where P(x)=x3+2x2+mx+nP(x) = x^3 + 2x^2 + mx + nP(x)=x3+2x2+mx+n, g(x)=(x−1)(x−3)g(x) = (x - 1)(x - 3)g(x)=(x−1)(x−3), and s(x)=3x+2s(x) = 3x + 2s(x)=3x+2. Substitute x=1x = 1x=1 P(1)=(1−1)(1−3)h(1)+(3(1)+2)P(1) = (1 - 1)(1 - 3)h(1) + (3(1) + 2)P(1)=(1−1)(1−3)h(1)+(3(1)+2)13+2(1)2+m(1)+n=0+51^3 + 2(1)^2 + m(1) + n = 0 + 513+2(1)2+m(1)+n=0+51+2+m+n=51 + 2 + m + n = 51+2+m+n=5m+n=2…(1)m + n = 2 \quad \ldots (1)m+n=2…(1) Substitute x=3x = 3x=3 P(3)=(3−1)(3−3)h(3)+(3(3)+2)P(3) = (3 - 1)(3 - 3)h(3) + (3(3) + 2)P(3)=(3−1)(3−3)h(3)+(3(3)+2)33+2(3)2+m(3)+n=0+113^3 + 2(3)^2 + m(3) + n = 0 + 1133+2(3)2+m(3)+n=0+1127+18+3m+n=1127 + 18 + 3m + n = 1127+18+3m+n=1145+3m+n=1145 + 3m + n = 1145+3m+n=113m+n=−34…(2)3m + n = -34 \quad \ldots (2)3m+n=−34…(2) Eliminate both equations 3m+n=−34…(2)m+n=2…(1)(2)−(1):(3m+n)−(m+n)=−34−2−2m=36m=−18\begin{aligned} 3m + n &= -34 \quad \ldots (2) \\ m + n &= 2 \quad \ldots (1) \\ \hline (2) - (1): \quad (3m + n) - (m + n) &= -34 - 2 \\ -2m &= 36 \\ m &= -18 \end{aligned}3m+nm+n(2)−(1):(3m+n)−(m+n)−2mm=−34…(2)=2…(1)=−34−2=36=−18 Substitute into equation (1) m+n=2m + n = 2m+n=2−18+n=2-18 + n = 2−18+n=2n=20n = 20n=20
