Question

WILL MEDALL & FANN

The area of a rectangular painting is given by the trinomial x^2 - 5x - 36. What are the possible dimensions of the painting? Use factoring.

a. (x+9) and (x-4)
b. (x+9) and (x+4)
c. (x-9) and (x-4)
d. (x-9) and (x+4)

Answer 1

@BoraSemiz

Answer 2

It just wants you to factor the trinomial given

Answer 3

What tow numbers multiply to be -36 and also add up to be -5

Answer 4

two*

Answer 5

-6 x 6?

Answer 6

= -36 right?

Answer 7

but I don't think -6+6 is -5

Answer 8

what about
-36 x 1
-5 x 1

Answer 9

you know you are looking for 2 numbers that when you multiply give you -36 AND also when you those same numbers they give you -5 right?

Answer 10

for example
what two numbers multiply to be -54 and add up to be 3?
well -6* 9=-54 and -6+9=3
so the two numbers I'm looking for in my example is -6 and 9

Answer 11

oh okay..

Answer 12

idk two numbers that equal -36 and -5....

Answer 13

no no...

Answer 14

you are looking for two numbers that multiply to be -36 and add up to be -5?

Answer 15

ohhhhhhhhhhhhhhhhh okay..

Answer 16

think about two numbers that multiply to be -36?
-1(36)=-36
1(-36)=-36
2(-18)=-36
-2(18)=-36
3(-12)=-36
-3(12)=-36
4(-9)=-36
-9(4)=-36

so you couldn't tell me out of these possibilities which of the pair of factors can you add that give you -5?

Answer 17

and we already discussed -6(6)
-6(6)=-36 but -6+6 isn't -5

Answer 18

4+-9

Answer 19

yes

Answer 20

so going back to my example
let me extend my example:
factor
\[x^2+3x-54 \\ \text{ we found two factors of -54 that have product -54 and sum 3 } \\ \text{ those numbers were } -6 \text{ and } 9 \\ \text{ the factored form of } x^2+3x-54 \text{ is } (x-6)(x+9)\]

Answer 21

so the factored form for x^2-5x-36 is (x+4) and (x-9)

Answer 22

well the factors of x^2-5x-36 is (x+4) and (x-9)
the factored form of x^2-5x-36 is (x+4)(x-9)

Answer 23

oh okay i get it!!... thxx so much for ur help!! :)

Answer 24

np
if you get stuck at factoring things in the form x^2+bx+c
you might want to try listing out the possibilities of c
(hopefully c won't be too large because you know that can get pretty lengthy)
and what I mean is you want two find two numbers that multiply to be c and also add up to be b
so those numbers are m and n
you are looking for m and n such that
c=m*n
and
b=m+n

so then you can write x^2+bx+c as (x+m)(x+n)
multiplying (x+m)(x+n) out we have
x^2+(m+n)x+mn
as you see it is true comparing it to
x^2+bx+c
we need mn to be c and m+n to be b

Answer 25

okaay.. thx again.. this really helped!! :)