Respuesta :
First we multiply 2r*r and it gives us [tex] 2r^{2} [/tex].
Then -5r, + 20r = 15r
Then -5*10 gives us -50.
So the answer is: [tex] 2r^{2}+15r-50 [/tex]
Then -5r, + 20r = 15r
Then -5*10 gives us -50.
So the answer is: [tex] 2r^{2}+15r-50 [/tex]
We'll use the FOIL method to multiply the factors.
FOIL
First, Outside, Inside, Last
Multiply the first terms, then the outside terms, then the inside terms, then the last terms.
First
(2r – 5)(r + 10)
[tex]2r \times r = 2r^2[/tex]
Outside
(2r – 5)(r + 10)
[tex]2r \times 10 = 20r[/tex]
Inside
(2r – 5)(r + 10)
[tex]-5 \times r = -5r[/tex]
Last
(2r – 5)(r + 10)
[tex]-5 \times 10 = -50[/tex]
The trinomial should now look like this:
[tex]2r^2 + 20r - 5r - 50[/tex]
Combine like terms:
[tex]20r - 5r = 15r[/tex]
The following trinomial will be your answer:
[tex]\boxed{2r^2 + 15r - 50}[/tex]
FOIL
First, Outside, Inside, Last
Multiply the first terms, then the outside terms, then the inside terms, then the last terms.
First
(2r – 5)(r + 10)
[tex]2r \times r = 2r^2[/tex]
Outside
(2r – 5)(r + 10)
[tex]2r \times 10 = 20r[/tex]
Inside
(2r – 5)(r + 10)
[tex]-5 \times r = -5r[/tex]
Last
(2r – 5)(r + 10)
[tex]-5 \times 10 = -50[/tex]
The trinomial should now look like this:
[tex]2r^2 + 20r - 5r - 50[/tex]
Combine like terms:
[tex]20r - 5r = 15r[/tex]
The following trinomial will be your answer:
[tex]\boxed{2r^2 + 15r - 50}[/tex]