Are there constants P and Q such that X4 + PX2 + Q can be divisible by x2 + 2x + 5? If it exists, find out the value of P and Q, otherwise, explain the reason

Are there constants P and Q such that X4 + PX2 + Q can be divisible by x2 + 2x + 5? If it exists, find out the value of P and Q, otherwise, explain the reason

If there exists, then it is shown that X4 + PX2 + Q can be divided by x 2 + 2x + 5, and another factor is x2 + MX + N, and we can set another factor is x 2 + MX + N, {(x2 + 2x + 5) (x2 + 2x + 5) (x2 + 2x + 2x + 5) (x2 + MX2 + Q, that is, there is X4 + (M + 2) X3 + (M + 2 + 2m + 5) x 2 + (n + 2m + 2 + 2m + 5 + 2 + 2m + 5m) x + 5 N = n + 2m + 2m + 2 = 0n + 2m + 2m + 2m + 5 = P and 2n + 2m + 5m = 2n + 5 = 2n + 5 = 2n + 5m = 2n + 5 N = 05n = q, and 2n + 5m = 2n + 5m = 2n + 5m = 05n = 2n + 5m = 05n = 05n = 05n = 05n = q solve the above equations, and the above equations, we solve the seeking p=6，q=25．

The product of (x + PX + 8) (x-2x + Q) does not contain the cubic term of X and X, and the value of P and Q can be obtained

If the quadratic term in the product of the original formula is 8x & # 178; - 2px & # 178; + QX & # 178; = (8-2p + Q) x & # 178; if it does not contain the square term of X, then 8-2p + q = 0 (1) if the cubic term in the product of the original formula is PX & # 179; - 2x & # 179; = (P-2) x & # 179; if it does not contain the cubic term of X, then P-2 = 0, P = 2 is substituted into (1) to get q = - 4

Decompose the factor in the range of real number: x3-x2-3x + 3

x3-x2-3x+3，=（x3-x2）-（3x-3），=x2（x-1）-3（x-1），=（x-1）（x2-3），=（x-1）（x+3）（x-3）．

Factoring in real numbers: x2-3x-2=______ ．

Let x2-3x-2 = 0, then a = 1, B = - 3, C = - 2, ｜ x = 3 ± (− 3) 2 − 4 × 1 × (− 2) 2 × 1 = 3 ± 172, ｜ x2-3x-2 = (x − 3 + 172) (x − 3 − 172). So the answer is: (x − 3 + 172) (x − 3 − 172)