Question

$factorise - \\ {x}^{3} + {13x}^{2} + 32x + 20$
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Answer 1

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...

Answer 2

$\huge\tt{\bold{\underline{\underline{Question᎓}}}}$

$factorise - \\ {x}^{3} + {13x}^{2} + 32x + 20$

$\huge\tt{\bold{\underline{\underline{Answer᎓}}}}$

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$= > {x}^{3} + 13 {x}^{2} + 32x + 20$

Now,here we use hit and trial method by substituting values to make them 0

Now,putting x=-1 in the given polynomial

$= > {( - 1)}^{3} + 13 {(1)}^{2} + 32( - 1) + 20$

$= > - 1 + 13 - 32 + 20$

$= > 12 - 12 = 0$

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

$= > {x}^{2} + 12x + 20 = 0$

$= > {x}^{2} + 10x + 2x + 20 = 0$

$= > x(x + 10) + 2(x + 10) = 0$

$= > (x + 2)(x + 10) = 0$

Therefore,(x+1)(x+2)&(x+10) are the factors of the given polynomial

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