Exercise 2.96. a. Implement the procedure pseudoremainder-terms, which is just like remainder-terms except that it multiplies the dividend by the integerizing factor described above before calling div-terms. Modify gcd-terms to use pseudoremainder-terms, and verify that greatest-common-divisor now produces an answer with integer coefficients on the example in exercise 2.95.
b. The GCD now has integer coefficients, but they are larger than those of P1. Modify gcd-terms so that it removes common factors from the coefficients of the answer by dividing all the coefficients by their (integer) greatest common divisor.
使用 pseudoremainder-terms
(define (pseudoremainder-terms L1 L2)
(define (integerize factor L)
(map
(lambda (x)
(make-term (order x) (mul factor (coeff x)))) L))
(let* ((factor (expt (coeff (first-term L2))
(+ 1
(order (first-term L1))
(- 0 (order (first-term L2))))))
(result (div-terms (integerize factor L1) L2)))
(cadr result)))
(define (gcd-terms a b)
(if (empty-termlist? b)
a
(gcd-terms b (pseudoremainder-terms a b))))
这个方法相当于在除数的位置乘以一个数字（被除数最高幂指数的系数的n次方），有了这个因子，在作除法的时候就可以实现系数的整除。
化简最终结果
在上面的结果中，我们已经得到了整数的系数，但系数是可以再化简的–再化简之后就变成了 p1，即最好的结果。
我们只需要把 gcd-terms 写成这样即可：
(define (gcd-terms a b)
(define (simplify terms)
(let ((gcd-fator (apply gcd (map (lambda (x) (coeff x)) terms))))
(map (lambda (term)
(make-term (order term) (div (coeff term) gcd-fator)))
terms)))
(if (empty-termlist? b)
(simplify a)
(gcd-terms b (pseudoremainder-terms a b))))
这个过程会计算所有系数的 gcd 值，然后每个系数除以这个 gcd 值就是最终解。
最后我们的测试如下：
(load "../testframe.scm")
(define p1 (make-polynomial 'x '((2 1) (1 -2) (0 1))))
(define p2 (make-polynomial 'x '((2 11) (0 7))))
(define p3 (make-polynomial 'x '((1 13) (0 5))))
(define q1 (mul p1 p2))
(define q2 (mul p1 p3))
(assertequal? (greatest-common-divisor q1 q2)
(make-polynomial 'x '((2 1) (1 -2) (0 1))))