Answer:
(x + [tex]\frac{1}{2}[/tex] + [tex]\frac{\sqrt{41} }{2}[/tex] )(x + [tex]\frac{1}{2}[/tex] - [tex]\frac{\sqrt{41} }{2}[/tex] )
Step-by-step explanation:
solve the polynomial for x then express in product of factor form
Solve
x² + x - 10 = 0 ( add 10 to both sides )
x² + x = 10
Using the method of completing the square
add ( half the coefficient of the x- term )² to both sides
x² + 2([tex]\frac{1}{2}[/tex] )x + [tex]\frac{1}{4}[/tex] = 10 + [tex]\frac{1}{4}[/tex]
(x + [tex]\frac{1}{2}[/tex] )² = [tex]\frac{41}{4}[/tex] ( take the square root of both sides )
x + [tex]\frac{1}{2}[/tex] = ± [tex]\sqrt{\frac{41}{4} }[/tex] = ± [tex]\frac{\sqrt{41} }{2}[/tex] ( subtract [tex]\frac{1}{2}[/tex] from both sides )
x = - [tex]\frac{1}{2}[/tex] ± [tex]\frac{\sqrt{41} }{2}[/tex]
Thus corresponding factors are
(x - (- [tex]\frac{1}{2}[/tex] - [tex]\frac{\sqrt{41} }{2}[/tex] ) ) and (x - (- [tex]\frac{1}{2}[/tex] + [tex]\frac{\sqrt{41} }{2}[/tex] ), that is
(x + [tex]\frac{1}{2}[/tex] + [tex]\frac{\sqrt{41} }{2}[/tex] ) and ( x + [tex]\frac{1}{2}[/tex] - [tex]\frac{\sqrt{41} }{2}[/tex] )
Thus
x² + x - 10 = ( x + [tex]\frac{1}{2}[/tex] + [tex]\frac{\sqrt{41} }{2}[/tex] )(x + [tex]\frac{1}{2}[/tex] - [tex]\frac{\sqrt{41} }{2}[/tex] ) ← in factored form
