Respuesta :
[tex]5 x^{3} - x^{2} + 5x - 1 [/tex]
Group like terms
[tex]( 5x^{2} - x^{2} ) + (5x - 1)[/tex]
Find the common factor of each term and simplify
[tex] x^{2} (5x - 1) + 1 (5x - 1) [/tex]
[tex]( x^{2} + 1) (5x - 1)[/tex]
Additional :-
Find the roots of each of the terms:
(i) [tex]( x^{2} + 1)[/tex]
- using the quadratic equation ([tex]x = \frac{-b (+ or -) \sqrt{ b^{2} - 4ac } }{2a} [/tex])
- [tex]\frac{-0 + \sqrt{0^{2 - 4(1)(1)} } }{2}[/tex] OR [tex]\frac{-0 - \sqrt{0^{2 - 4(1)(1)} } }{2}[/tex]
- Since the discriminant is negative ([tex] \sqrt{-4} [/tex]) [a negative number cannot be rooted] then this equation has no real roots (imaginary roots)
(ii) [tex]5x - 1[/tex]
- Simply Solve for x
5x - 1 = 0
5x = 1
x = [tex] \frac{1}{5} [/tex]
Thus the only solution this has or the only root this expression has is [tex] \frac{1}{5} [/tex] OR 0.2
Group like terms
[tex]( 5x^{2} - x^{2} ) + (5x - 1)[/tex]
Find the common factor of each term and simplify
[tex] x^{2} (5x - 1) + 1 (5x - 1) [/tex]
[tex]( x^{2} + 1) (5x - 1)[/tex]
Additional :-
Find the roots of each of the terms:
(i) [tex]( x^{2} + 1)[/tex]
- using the quadratic equation ([tex]x = \frac{-b (+ or -) \sqrt{ b^{2} - 4ac } }{2a} [/tex])
- [tex]\frac{-0 + \sqrt{0^{2 - 4(1)(1)} } }{2}[/tex] OR [tex]\frac{-0 - \sqrt{0^{2 - 4(1)(1)} } }{2}[/tex]
- Since the discriminant is negative ([tex] \sqrt{-4} [/tex]) [a negative number cannot be rooted] then this equation has no real roots (imaginary roots)
(ii) [tex]5x - 1[/tex]
- Simply Solve for x
5x - 1 = 0
5x = 1
x = [tex] \frac{1}{5} [/tex]
Thus the only solution this has or the only root this expression has is [tex] \frac{1}{5} [/tex] OR 0.2