Question

用十字相乘法分解因式:a^2b^2+16ab+39

（2）15x^2n+7x^ny^n+1-4y2n+2
（3）（x^2+3x）^2-22（x^2+3x）+72

Answer 1

(1) a^2b^2+16ab+39
1 13
x
1 3
=(ab+13)(ab+3)
(2) 15x^2n+7x^ny^n+1-4y2n+2
5 -4
x
3 1
=(5x^n -4y^n+1)(3x^n+ y^n+1)
(3) （x^2+3x）^2-22（x^2+3x）+72
1 -4
x
1 -18
=[（x^2+3x）-4] * [（x^2+3x）-18]
1 4 1 6
x x
1 -1 1 -3
=(x+4)(x-1)(x+6)(x-3)