Question

3. Use factor theorem to show that (x-2) is a factor of the polynomial x^3-x-6 and (x+2)(x-1) is a factor of
x^4+ x^3+2x^2+4x-8.

Answer 1

using factor theorem

x-2=0

=x=2

Put value of x = 2

${2}^{3} - 2 - 6 \\ = 0 \\ therefore \: 2 \: is \: the \: root \: of \: p(x). \\ thus \: x - 2 \: is \: a \: factor \: of \: p(x)$

II

Using factor theorem

(x+2)(x-1)=0

$using \: identity \: (x + a)(x + b) = {x}^{2} + (a + b)x + ab \\ \\ = {x}^{2} + (2 - 1)x - 2 = 0 \\ {x}^{2} + x - 2 = 0 \\ {x}^{2} + x = 2 \\ x(x + 1) = 2$

after this it is a little tough to express it using these formulae writing as its hard to type each and every thing in detail and as it is midnight in india I can't just express it on a piece of paper but if you make some effort you will be successful

Answer 2

Hola mate

Plz refer to the given attachment and do the question by understanding the concept from the attachment...