Question

(x-1)(x-2)(x-5)(x-7)-65​

Answer 1

Answer:

Step-by-step explanation:

We need to factorize:

(x−1)(x−3)(x−5)(x−7)−64

(x−1)(x−3)(x−5)(x−7)−64

We can, by the rational root theorem, see that there are no roots of this polynomial.Next observation is that 64=(8)264=(8)2. So this means that if the first part of the polynomial is a square,we can rewrite the whole polynomial as the difference of two squares. But it turns out that the first part of the polynomial is not a square. However,we can note that,

(x−1)(x−7)=(x2)−8x+7

(x−1)(x−7)=(x2)−8x+7

(x−3)(x−5)=(x2)−8x+15

(x−3)(x−5)=(x2)−8x+15

Therefore,letting (x2)−8x+7=p(x2)−8x+7=p,we can rewrite the given polynomial as

p(p+8)−64=p^2+8p−64

now it is factorisable..

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