Question

All the real zeros of the given polynomial are integers. Find the zeros.

Answer 1

Answer:

x = - 7 , x = 2 , x = 1

Step-by-step explanation:

note that

p(1) = 1³ + 4(1)² - 19(1) + 14 = 1 + 4 - 19 + 14 = 0

since p(1) = 0 , then x = 1 is a zero

using synthetic division

1 | 1  4  - 19    14

    ↓  1     5   - 14  

   ------------------------

     1  5    - 14   0

quotient is x² + 5x - 14

equating quotient to zero

x² + 5x - 14 = 0

(x + 7)(x - 2) = 0

equate each factor to zero and solve for x

x + 7 = 0 ⇒ x = - 7

x - 2 = 0 ⇒ x = 2

then zeros are x = - 7, x = 2 , x = 1

Answer 2

Answer:

$x = 1\\\\x =2\\\\x = -7$

Step-by-step explanation:

$~~~~~~~x^3 +4x^2 -19x +14=0\\\\\implies x^3- x^2+5x^2 -5x -14x+14=0\\\\\implies x^2(x-1) +5x (x-1)-14 (x-1)=0\\\\\implies (x-1)(x^2 +5x -14) = 0\\\\\implies (x-1) (x^2 +7x -2x -14) = 0\\\\\implies (x-1) \textbf{[}x(x+7) -2(x+7) \textvf{]}=0\\\\\implies (x-1)(x-2)(x+7)=0\\\\\implies x = 1,~ x =2,~ x = -7$