X4 + AX3 + x2 + BX + 1 has a factor x2-2x + 1 to find a, B and decompose the factor

X4 + AX3 + x2 + BX + 1 has a factor x2-2x + 1 to find a, B and decompose the factor

∵ X4 + AX3 + x2 + BX + 1 has a factor X ^ 2-2x + 1 = (x-1) ^ 2
Let X4 + AX3 + x2 + BX + 1 = (x ^ 2 + MX + n) (x ^ 2-2x + 1)
The constant term n * 1 = 1, n = 1
Right quadratic term: MX * (- 2x) + NX ^ 2 + x ^ 2 = (- 2m + 2) x ^ 2
Left quadratic: x ^ 2
∴ -2m+2=1,m=1/2
∴x4+ax3+x2+bx+1=(x^2+1/2x+1)(x^2-2x+1)
Right third order item: 1 / 2x * x ^ 2-2x * x ^ 2 = - 3 / 2x ^ 3
Left third power: ax ^ 3
Right 1 time item: 1 / 2x-2x = - 3 / 2x
Left 1st order: BX
∴ a=-3/2,b=-3/2
∴x⁴-3/2x³+x²-3/2x+1=(x^2+1/2x+1)(x-1)²

Decomposition factor: 4 (3x2-x-1) (x2 + 2x-3) - (4x2 + x-4) 2

4(3x^2-x-1)(x^2+2x-3)-(4x^2+x-4)2
=4(3x^4+5x^3-12x^2+x+3)-(16x^4+8x^3-31x^2-8x+16)
=-4x^4+12x^3-17x^2+12x-4
=-x^2(4x^2-12x+9)-4(2x^2-3x)-4
=-(2x^2-3x)^2-4(2x^2-3x)-4
=-(2x^2-3x+2)^2

Given that x2 + 2x + 5 is a factor of X4 + AX2 + B, find the value of a + B

Let X4 + AX2 + B = (x2 + 2x + 5) (x2 + MX + n) = X4 + (2 + m) X3 + (2m + N + 5) x2 + (5m + 2n) x + 5N. By comparing the coefficients of corresponding terms, we get 2 + M = 02m + N + 5 = a5m + 2n = 05n = B. the solutions are m = - 2, n = 5, a = 6, B = 25 ∥ a + B = 31