Question

Multiplying binomial by a trinomial now.
Equation in comments

Answer 1

(4x^3-y^4)(3x+4x^8-9)

Answer 2

@ganeshie8

Answer 3

same method, we distribute term by term

Answer 4

\((4x^3-y^4)(3x+4x^8-9) \)

Answer 5

\((\color{green}{4x^3}-y^4)(\color{green}{3x+4x^8-9}) \)

Answer 6

I can't see equation using the buttons.

Answer 7

first distribute 3x^3, then \(-y^4\)

Answer 8

ohh i think there is some bug, its not working properly i heard mods were saying equation editor not working properly

Answer 9

(4x^3-y^4)(3x+4x^8-9) is the equation.

Answer 10

hmm didnt get u

Answer 11

you put (3x^3) its (4x^3)

Answer 12

oh gotcha :)

Answer 13

lol that was a typpo.. nothing to do wid editor hehe

\( (\color{green}{4x^3}-y^4)(\color{green}{3x+4x^8-9}) \)

Answer 14

okay.

Answer 15

yea distribute \(4x^3\) first

Answer 16

okay.

Answer 17

is it:
(12x^3+16x^11y^5-36x^3)?

Answer 18

\((\color{green}{4x^3}-y^4)(\color{green}{3x+4x^8-9}) \)

\(\color{green}{4x^3}(\color{green}{3x+4x^8-9}) - y^4(3x+4x^8-9) \)

Answer 19

I can't see the equations used from the buttons! please just type them! im totally confunsed. :/

Answer 20

oh you cant see the latex thats frustrating

Answer 21

all i see is \((\color{})(\color{}{}) \) but fulled with the equation

Answer 22

(4x^3 - y^4)(3x+4x^8-9)

4x^3(3x+4x^8-9) - y^4(3x+4x^8-9)

Answer 23

anyhoo, i have (12x^3+16x^11y^5-36x^3)

Answer 24

is that better

Answer 25

yea, thank you i can read it now! :)

Answer 26

4x^3(3x+4x^8-9) - y^4(3x+4x^8-9)

(12x^4 + 16x^11 - 36x^3) - (3xy^4 + 4x^8y^4 - 9y^4)

Answer 27

i have to condense my paper, give me a second.

Answer 28

(12x^4 + 16x^11 - 36x^3- 3xy^4 - 4x^8y^4 + 9y^4)

Answer 29

ok if u see any like terms, just add them. i think somthing went wrong in ur answer.. . check :)

Answer 30

for which part of the equation though?

Answer 31

i dont see any like terms. so thats it i hope

Answer 32

(12x^4 + 16x^11 - 36x^3- 3xy^4 - 4x^8y^4 + 9y^4)

Answer 33

12x^4 + 16x^11- 36x^3, 3xy^4 - 4x^8y^. are all like terms

Answer 34

why do u think they are like terms

Answer 35

they are all the same i.e. (x)^4,(x)^11,(-x)^3; (xy)^4,(x)^8(y)^4

Answer 36

@ganeshie8

Answer 37

(x)^4 and (x)^11 are not like terms

Answer 38

to be called as like terms, terms should satisfy two things-
1) variable letters, for example x, y
2) exponent of variables

Answer 39

terms listed by you are satisfying first condition.
but they are failing second condition

Answer 40

if u see,

(x)^4,(x)^11


first one has an exponent of 4, and second one has an exponent of 11

so they are not like terms. we cant add them

Answer 41

oh but they are. they are both (x)s. so by that, all you would do is att the exponets and the numerals.

Answer 42

for addition, BOTH variable(x)s and exponents(4, 11) MUST be equal
for multiplication, variables must be equal.