Question

Please friends solve this.

Answer 1

1)
$p(x) = x {}^{3} + 2x {}^{2} + 3x + 2$

put -1 in eq.

$p( - 1) = ( - 1) {}^{3} + 2( - 1) {}^{2} + 3( - 1) + 2$


$= - 1 + 2 - 3 + 2 = 0$

Hence -1 satisfy thee equation so (x + 1) is a factor.

Now, f (x) ÷ g (x)

$(x {}^{3} + 2x {}^{2} + 3x + 2) \div (x + 1)$
$(x {}^{2} + x + 2)(x + 1)$

applying quadratic formula in
$x {}^{2} + x + 2$

a =1 , b = 1 , c = 2

$x = (- b + - \sqrt{b {}^{2} - 4ac} ) \div 2a$

$( - 1 + - \sqrt{(1) {}^{2} - 4(1)(2)} ) \div 2(1)$
$( - 1 + - \sqrt{ - 7} ) \div 2$


So roots are
$(x + (1 + \sqrt{ - 7}) \div 2)$

$(x + (1 - \sqrt{ - 7} ) \div 2)$

$(x + 1)$

2)
$p(x) = x {}^{3} - x{}^{2} - x + 1$

put 1 in eq
$(1) {}^{3} - (1) {}^{2} - 1 + 1$

Hence 1 satisfied eq. so (x - 1) is a factor.

$(x {}^{3} - x {}^{2} - x + 1) \div (x - 1)$

$(x {}^{2} - 1)(x - 1)$

$(x - 1)(x + 1)(x - 1)$


Thanks