Answered

Discover answers to your questions with Westonci.ca, the leading Q&A platform that connects you with knowledgeable experts. Our platform provides a seamless experience for finding precise answers from a network of experienced professionals. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.

All the real zeros of the given polynomial are integers. Find the zeros.
P(x) = x³ + 4x2 - 19x + 14
X =

Sagot :

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

loneguy

Answer:

[tex]x = 1\\\\x =2\\\\x = -7[/tex]

Step-by-step explanation:

[tex]~~~~~~~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[/tex]

Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.