Question

the zeros of the polynomial x2(it's x square)-3x-4 are. a)4,1. b)4,-1. c)-4,1. d)-4,-1​

Answer 1

p(x)=x²-3x-4

let p(x)=0

=>x²-3x-4=0

=>x²+x-4x-4=0

=>x(x+1)-4(x+1)=0

=>(x+1)(x-4)=0

:.x =4,-1

so option (b) is the answer

Answer 2

Answer

(⌒o⌒)(⌒o⌒)(⌒o⌒)(⌒o⌒)(⌒o⌒)(⌒o⌒)(⌒o⌒)(⌒o⌒)(⌒o⌒)(⌒o⌒)(⌒o⌒)(⌒o⌒)(⌒o⌒)(⌒o⌒)(⌒o⌒)(⌒o⌒)

$\pink{\huge{\underline{\underline{❤Answer❤}}}}$

$\bigstar\small \boxed { \sf \red{ \: the \: zeros \: of \: the \: polynomial \: {x}^{2} - 3x - 4 \: are = \: to \: find }}$

$\small \boxed { \sf \pink { options: (a) 4 and 1 }}$

$\small \boxed { \sf \pink {(b)4 \: and \: - 1}}$

$\small \boxed{ \sf \pink{(c) \: - 4 \: and \: 1}}$

$\small \boxed { \sf \pink{(d) \: - 4 \: and \: - 1}}$

$\underline { \sf{solution \: }}$

$\implies \displaystyle \sf{ {x}^{2} - 3x - 4 = 0}$

$\implies \displaystyle \sf{ {x}^{2} - 4x + x - 4 = 0}$

$\implies \displaystyle \sf{x(x - 4) + 1(x - 4) = 0}$

$\implies \displaystyle \sf{(x - 4) = 0 \: and \: (x + 1) = 0}$

$\implies \displaystyle \sf{x = 4 \: and \: x = - 1}$

$\pink \bigstar \displaystyle \boxed{ \sf \green{ \: option(b) \: 4 \: and \:- 1 \: is \: correct }}$