[tex]factorise - \\ {x}^{3} + {13x}^{2} + 32x + 20[/tex]
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[tex]factorise - \\ {x}^{3} + {13x}^{2} + 32x + 20[/tex]
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Answer:
x³+13x²+32x+20
by trial n error method
x(-1) = x³+13x²+32x+20
(-1)³+13(-1)²+32(-1)+20
-1+13-32+20
0
therefore, (x+1) is a factor
now, divided x³+13x²+32x+20 by (x+1)
= x²+12x+20
x²+12x+20
x²+10x+2x+20
x(x+10)+2(x+10)
(x+2)(x+10)
All the factors are :- (x+1)(x+2)(x+10)
hope it helps you...
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Now,here we use hit and trial method by substituting values to make them 0
Now,putting x=-1 in the given polynomial
Hence,x+1 is a factor of the given polynomial.
we have to find another two roots ;for this divide the given polynomial by (x+1)
=>see in the picture for dividing process :-
After dividing we get x^2+ 12x+20(quotient) now we have to factorise the quotient to find other roots
Therefore,(x+1)(x+2)&(x+10) are the factors of the given polynomial
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