V^2+10v+15 Trinomal factoring where [a]=1

Question

please help!

Answer 1

Given trinomial is v^2+10v+15

To clarify of given problem i assume the leading coefficient is a = 1

Now we can factorize the given trinomial.

Compare it to quadratic form ax^2+bx+c

Here x = v, a = 1, b = 10, c = 15

v = [-b±√b^2-4ac]/2a

v = [-10±√(100-60)/2

v = [-10±√40]/2

v = [-10±√4*10]/2

v = [-10±2√10]/2

v = 2[-5±√10]/2

v = -5±√10

v = -5+√10 and v = -5-√10

Factoring of given trinomial is v^2+10v+15 = (v-(-5+√10))(v-(-5-√10)) =(v+5-√10)(v+5+√10)