factor the trinomial x^2+ 16x + 7

Question

need help with this

Answer 1

There is a way to do this, but it requires you to find the solutions when you set this trinomial equal to zero.

 

x² + 16x + 7 = 0

 

 

Use quadratic formula:

 

x = (-b ± √(b² - 4ac)) / 2a

 

where:

a = 1

b = 16

c = 7

 

Plugging in the values,

 

x = (-16 ± √(256 - 28)) / 2

x = (-16 ± √228) / 2

x = (-16 ± √4√57) / 2

x = (-16 ± 2√57) / 2

x = -8 ± √57

 

x = -8 - √57                   and                     x = -8 + √57

 

We use this fact to factor.

 

x² + 16x + 7 = (x + 8 + √57)(x + 8 - √57)

Answer 2

Use the quadratic formula to find the roots, and then obtain the factors from the roots.

 

x^2 +16x + 7 = 0

 

x1 = {-16 + √[(16)^2 - 4(1)(7)]} / 2 = -0.450166

 

x2 = {-16 - √[(16)^2 - 4(1)(7)]} / 2 = -15.549834

 

Then

 

x^2 +16x + 7 = (x - x1)(x - x2) = (x + 0.450166)(x + 15.549834)

 

Answer 3

Hi joyce

The trinomial is not factorable over the set of integers.You can use long division to divide the trinomial by multiple binomials, but the quotient or other factor would have a remainder.