Question

Medal and fanning !!!!!!!!!
thank you ^_^
1. x2 – 8x + 15

2. a2 – a – 20

3. a2 – 5a – 20

4. a2 + 12ab + 27b2

5. 2a2 + 30a + 100

Answer 1

1. x^2 - 8x + 15
x^2 - 5x - 3x + 15
x(x - 5) -3(x - 5)
(x - 3)(x - 5)

2. a^2 - a - 20
a^2 - 5a + 4a - 20
a(a - 5) 4(a - 5)
(a + 4)(a - 5)

3. a^2 - 5a - 20
Unfactorable.

4. a^2 + 12ab + 27b^2
a^2 + 9ab + 3ab + 27b^2
a(a + 9b) 3b(a + 9b)
(a + 3b)(a + 9b)

5. 2a^2 + 30a + 100
2a^2 + 20a + 10a + 100
2a(a + 10) 10(a + 10)
(2a + 10)(a + 10)

Answer 2

@Hero

Answer 3

From where did you learn this factoring method? It's the exact same kind I use.

Answer 4

By the way, a^2 - 5a - 20 is factorable

Answer 5

Flvs but i think two of them are wrong

Answer 6

Ohh can you show me how hero? thank you

Answer 7

For Number 5, the factoring steps are this:
2a^2 + 30a + 100
= 2(a^2 + 15a + 50)
= 2(a^2 + 10a + 5a + 50)
= 2((a(a + 10) + 5(a + 10))
= 2((a + 10)(a + 5))
= 2(a + 10)(a + 5)

Answer 8

Thanks thats what i thought too but please tell me which ones are wrong and which ones are right and if whatever ones i got wrong please show me the right way thanks hero

Answer 9

Technically you didn't get any of the wrong. I was just saying that number 3 is factorable.

Answer 10

Oh ok thanks but can you explain how its factorable because I don't know how

Answer 11

and do you think for number two the sign needs to be changed?

Answer 12

The sign for number 2 doesn't need to be changed because

\(-5 \times 4 = -20\)
\(-5 + 4 = -1\)

So you found the right two numbers

Answer 13

Number 3 would take a bit more effort to explain.

Answer 14

Ok thank you but could you please help for number three

Answer 15

a^2 - 5a - 20


You need two numbers \(m\) and \(n\) that multiply to get -20 yet add to get -5

In other words

\(mn = -20\)
\(m + n = -5\)

It's a bit much to show here.

Answer 16

hows about 4*-5 equals =-20

Answer 17

-5 + 4 doesn't add up to -5

Answer 18

and 15-20=-5

Answer 19

Ohhh

Answer 20

m and n have to be the same

Answer 21

Then i do not think its possible

Answer 22

It's possible

Answer 23

Hmm

Answer 24

What I mean is m and n have to be consistent for both equations.

Answer 25

Like what I showed you above for number two.

Answer 26

Yeah i see but i don't know which numbers would work

Answer 27

Trust me, it's not worth knowing at your level right now.

Answer 28

Loll I just searched that one up every ones saying its prime

Answer 29

Some one says this It is prime and will not factor.

If you are looking for roots (zeros), then you can complete the square or use the quadratic formula.

Let's complete the square.

a^2 - 5a - 20 = 0
Add 20 to both sides:
a^2 - 5a = 20
Add (-5/2)^2 to both sides to complete the square:
a^2 - 5a + 25/4 = 20 + 25/4
Factor the left side, and simplify the right:
(a - 5/2)^2 = 105/4
Take the square root of both sides, but be careful not to lose the negative root:
a - 5/2 = (±√105) / 2
Add 5/2 to both sides:
a = (5±√105)/2

Answer 30

what do you think about it @Hero

Answer 31

Of course you can find the values that way. But I was referring to the factoring method.

Answer 32

I'll show it to you later. I'm going to go take a nap.

Answer 33

Lol Ok have a good nap

Answer 34

@SOS101