Question

draw the graph of the given polynomial and find the zeroes. justify the answer. p(x)=x^2+3x-4​

Answer 1

Answer:

-4,2 are the zeroes of the polynomial

plz followe me

Answer 2

Answer:

$x^2+3x-4$

$x^2+4x-1x-4=0\\x(x+4) -1(x+4) = 0\\(x+4) (x-1)=0$

($x$ + 4) = 0 or ($x$ -1) = 0

($x$  = -4) or ($x$  = 1)

Rough:-

$y$ = $x^2+3x-4$

⇒ p(1) = (1)² + 3(1) - 4

= 1 + 3 - 4

= 4 - 4 = 0 [0,1]

⇒ p(2) = (2)² + 3(2) - 4

= 4 + 6 - 4

= 10 - 4 = 6 [2,6]

⇒ p(3) = (3)² + 3(3) - 4

= 9 + 9 - 4

= 18 - 4 = 14 [3,14]

⇒ p(4) = (4)² + 3(4) - 4

= 16 + 12 - 4

= 28 - 4 = 24 [4,24]

⇒ p(0) = (0)² + 3(0) - 4

= 0 + 0 - 4 = 4 [0,4]

⇒ p(-1) = (-1)² + 3(-1) - 4

= 1 - 3 - 3

= 1 - 7 = -6 [-1,-6]

⇒ p(-2) = (-2)² + 3(-2) - 4

= 4 - 6 - 4

= 4 - 10 = -6 [-2,-6]

⇒ p(-3) = (-3)² + 3(-3) - 4

= 9 - 9 - 4 = -4 [-3,-4]

⇒ p(-4) = (-4)² + 3(-4) - 4

= 16 - 12 - 4

= 16 - 16 = 0 [-4,0]

⇒ p(-5) = (-5)² + 3(-5) - 4

= 25 - 15 - 4

= 25 - 19 = 6 [-5,6]