Can someone help me with this problem?

Question

not just the answer i mean explain the whole process and everything? its only a practice problem but i wanna learn to do it.

(4a-3)(5a+1)

Answer 1

You want to use the FOIL method, which means First, Outer, Inner, and Last:
F(4a*5a)+O(4a*1)+I(-3*5a)+L(-3*1)
20a^2+4a-15a-3
20a^2-11a-3

I hope this helps!

Answer 2

You don't say what you need to do with it....?

If it's equal to zero:

(4a-3)(5a+1) = 0

Then either 4a - 3 = 0 (so a = 3/4) or 5a + 1 = 0 (so a = -1/5).

If you're supposed to convert it to a polynomial, then just do FOIL

First: 4a(5a) = 20a²
Outer: 4a(1) = 4a
Inner: -3(5a) = -15a
Last: -3(1) = -3

So the result would be 20a² + 4a - 15a - 3 = 20a² - 11a - 3.

Answer 3

Binomial*Binomial scenario is done using FOIL.

(a+b)(c+d)=ac+ad+bc+bd

choose a particular one of the brackets and multiply each seperate variable within that bracket with each other variable in the other bracket

in your case a=4a b=-3 c=5a d=1

ac=(4a)(5a) +ad=(4a)(1) +bc=(-3)(5a) +bd=(-3)(1)

= 20a^2+4a-15a-3 simplify

= 20a^2 -11a -3

Answer 4

(4a-3)(5a+1) use FOIL
First: 20a^2
Outside: 4a
Inside: -15a
Last: -3
add them up
20a^2-11a--3

Answer 5

20a squre + 4a -15a -3
20 a squre -11-3
20 a squre -11-3
+3 +3
20 a squre -8=0
( a + 2 ) ( a _ 10 )=0
a+2=0 or a-10=0
-2 -2 +10 +10
a=-2 or a =10

i use dthe distributive property and foil

Answer 6

distributing
4a(5a+1)-3(5a+1)
=20a^2+4a-15a-3
=20a^2-11a-3

Answer 7

Please be specific on what needs to be done.

(4a)(5a) + (4a)(1) - (3)(5a) - (3)(1)

20a^2 + 4a - 15a - 3

20a^2 - 11a - 3

Answer 8

20a(a being squared) + 11a-3