Respuesta :
To factor the trinomial, you can use the ac method where you multiply the first and last terms together (6 and 9) and add up to 15.
In this case ac= 54
2 numbers that multiply to 54 and add up to 15 would be 6 and 9 so place these numbers where the 15 would be:
[tex] 6p^{2} +9p + 6p+9[/tex]
Then group terms to simplify:
[tex] (6p^{2} +9p)(6p+9) [/tex]
Simplify any like terms:
[tex] 3p(2p+3) 3(2p+3)[/tex]
And group together again:
[tex](3p+3)(2p+3)[/tex]
You can simplify a 3 out of the first expression one more time to get:
[tex]3(p+1)(2p+3)[/tex]
Which would be the factored form. I hope this helps :)
In this case ac= 54
2 numbers that multiply to 54 and add up to 15 would be 6 and 9 so place these numbers where the 15 would be:
[tex] 6p^{2} +9p + 6p+9[/tex]
Then group terms to simplify:
[tex] (6p^{2} +9p)(6p+9) [/tex]
Simplify any like terms:
[tex] 3p(2p+3) 3(2p+3)[/tex]
And group together again:
[tex](3p+3)(2p+3)[/tex]
You can simplify a 3 out of the first expression one more time to get:
[tex]3(p+1)(2p+3)[/tex]
Which would be the factored form. I hope this helps :)
actor using the ac method
for
ax²+bx+c when a≠1
multiply a and c and remember their product
find what 2 numbers multiply to get that product and add to be
split b into those 2 numbers
factor
so
first factor the 3
3(2p²+5p+3)
multiply a and c
2*3=6
wat 2 numbers multiply to get 6 and add to 5
oh ya, 2 and 3
3(2p²+2p+3p+3)
gropu and factor
3((2p²+2p)+(3p+3))
3(2p(p+1)+3(p+1))
3(2p+3)(p+1) is factored
for
ax²+bx+c when a≠1
multiply a and c and remember their product
find what 2 numbers multiply to get that product and add to be
split b into those 2 numbers
factor
so
first factor the 3
3(2p²+5p+3)
multiply a and c
2*3=6
wat 2 numbers multiply to get 6 and add to 5
oh ya, 2 and 3
3(2p²+2p+3p+3)
gropu and factor
3((2p²+2p)+(3p+3))
3(2p(p+1)+3(p+1))
3(2p+3)(p+1) is factored