factorise the following polynomial
x³-19x-30
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Given, f(x)= x3 – 19x – 30
In f(x) putting x=±1, ±2, ±3, we see for which value of x, f(x)=0
f(−2)=(−2)3−19.(−2)+30=0
We observe that f(−2) = 0
From factor theorem, we can say, (x+2) is a factor of f(x)
x3 – 19x – 30 = x3 – 19x – 30
= x3+2x2−2x2−4x−15x −30
= x2(x+2)−2x(x+2)−15(x+2)
= (x+2)(x2−2x−15)
= (x+2)(x2−5x+3x−15)
= (x+2)( x(x−5) + 3(x−5) )
= (x+2)(x−5)(x+3)
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