how to fractorise using factor theorem??
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How do l factorise x3−3x2−9x−5x3−3x2−9x−5 by using factor theorem?
answer :
notice, x=−1x=−1 satisfies the cubic equation: x3−3x2−9x−5=0x3−3x2−9x−5=0 hence from factor theorem, x+1x+1 is a factor of x3−3x2−9x−5x3−3x2−9x−5
∴x3−3x2−9x−5∴x3−3x2−9x−5
=x2(x+1)−4x(x+1)−5(x+1)=x2(x+1)−4x(x+1)−5(x+1)
=(x+1)(x2−4x−5)=(x+1)(x2−4x−5)
=(x+1)(x2−5x+x−5)=(x+1)(x2−5x⏟+x−5⏟)
=(x+1)(x(x−5)+(x−5))=(x+1)(x(x−5)+(x−5))
=(x+1)(x−5)(x+1)=(x+1)(x−5)(x+1)
=(x−5)(x+1)2
answer :
notice, x=−1x=−1 satisfies the cubic equation: x3−3x2−9x−5=0x3−3x2−9x−5=0 hence from factor theorem, x+1x+1 is a factor of x3−3x2−9x−5x3−3x2−9x−5
∴x3−3x2−9x−5∴x3−3x2−9x−5
=x2(x+1)−4x(x+1)−5(x+1)=x2(x+1)−4x(x+1)−5(x+1)
=(x+1)(x2−4x−5)=(x+1)(x2−4x−5)
=(x+1)(x2−5x+x−5)=(x+1)(x2−5x⏟+x−5⏟)
=(x+1)(x(x−5)+(x−5))=(x+1)(x(x−5)+(x−5))
=(x+1)(x−5)(x+1)=(x+1)(x−5)(x+1)
=(x−5)(x+1)2
Anonymous:
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