(X-1)(x-3)(x-5)(x-7)-65 factorised
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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..
hope it helpsss
(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..
hope it helpsss
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