Question

Will give medal + fan to whoever can help me step-by-step on two questions. I really need to finish this by today so anyone please thanks in advanced (:
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Answer 1

yo

Answer 2

hey

Answer 3

sorry cant help but @johnweldon1993 probably can

Answer 4

@johnweldon1993 please help

Answer 5

thank you.

Answer 6

u r welcom

Answer 7

@Xmoses1 here ya go.

Answer 8

Um yeah this is not my specialty. Math is my worst area :\

Answer 9

same here.

Answer 10

Here i have someone who might know. He is really good in math @AkashdeepDeb

Answer 11

Thank you anyway (:

Answer 12

You are welcome i'm sorry that i couldn't help :\

Answer 13

Haha, thanks @Xmoses1 . :')

In the first question, what do they mean by 'this method'? :)

Answer 14

Its alright. and i think it means by using the same method used to factor the four-term polynomials which is "grouping"

Answer 15

I'll tell you the method they are talking about.

There is a method known as Middle-term splitting for factorizing quadratic polynomials.
Do you know what quadratic polynomials are?

Answer 16

yes; x^2+3x-40 is an example of a quadratic polynomial (this case trinomial) right?

Answer 17

Yes correct! :D
Is it urgent, or can I explain, middle-term splitting in 10 minutes?

Answer 18

well its long overdue and i have to call my teacher soon lol im sorry its kinda urgent

Answer 19

Okay, I'll explain.

MIDDLE TERM SPLITTING METHOD.

The standard form of a quadratic equation = ax2 + bx + c

Now the rule tells us that we must find 2 integers p and q

such that

p*q = c
p±q = b

For example:

x2 + 4x + 3 is a quadratic equation.

Here we need to find 2 integers p and q such that
p*q = 3
p+q = 4

Do you know of any such integers?

Answer 20

1 and 3? since 1x3=3 and 1+3=4

Answer 21

Excellent.

So

p = 1 , q = 3 or p = 3, q = 1 [DOES NOT MATTER. :P ]

So we can usually write our equation as

x2 + 4x + 3
= x2 + (1+3)x + 3
= x2 + 1x + 3x + 3
= x(x+1) + 3(x+1)
Now we have (x+1) common

so = (x+1)(x+3) [ab + cb = b(a+c) ]

Getting this? :)

Answer 22

I understood everything except how did you get x(x+1)+3(x+1) kinda confused on that.

Answer 23

See

we got p = 1 and q = 3 yeah?

And we know our standard equation.

x2 + (p+q)x + pq
When we put the values

x2 + (1+3)x + 3

[Our main intention is to split it into 4 terms to get it factorized]

Opening the parenthesis we get

x2 + 1x + 3x + 3

Now we re-group them correctly, so that they may be factorized.

(x2 + 1x) + (3x+3)

[x(x+1)] + [3(x+1)]

Now we have (x+1) common, so we can group them together

[x+1](x+3)

Answer 24

Ohh okay

Answer 25

Try this:

x2 - x - 6 [a=1 , b = -1, c = -6]

Here we need 2 integers

which give

p+q = -1
pq = -6

What do you think the values of p and q are?

Answer 26

im sorry i dont really know /:

Answer 27

No probs. :)

-6 can have it's multiples as

p * q
-3 * 2
-2 * 3
-1 * 6
-6 * 1

But we also need the integers to be
p+q = -1

when p = -2 , q = 3
p + q = 1
NOT CORRECT.

when p = -1 , q = 6
p+q = 5
NOT CORRECT.

when p = -6 and q = 1
p+q = -5
NOT CORRECT.

Thus, only when
p = -3, q = 2
p+q = -3 + 2 = -1 !! :D

So we can concur that p = -3 and q = 2

because

p+q = -1
p*q = -6

So

x2 - x - 6 can be written as

x2 + (p+q)x + pq

x2 + (-3 + 2)x + (-3 * 2) = x2 + (-3 + 2)x -6

x2 - 3x + 2x - 6
x(x-3) + 2(x-3)

TAKING (x-3) common we get, (x+2)(x-3) [the factorized form]

Getting it?

Answer 28

Yes i am thank you.

Answer 29

Wanna try one more example?

x2 - 8x + 16

p+q = -8
pq = +16

what can p and q be?

Answer 30

HINT:

Think about the integer factors of 16.
If you do not get it, ask, I'll explain again, no issues.

Answer 31

lol im sorry i have never been so good when there is a negative number involved

Answer 32

No problem.

Do you know negative number rules?

Like

-4 * -2 = 8
4 * 2 = 8
4 * -2 = -8
-4 * 2 = -8

-4 + 2 = -2
-4 - (+2) = -6

-4 + (-2) = -6
-4 - (-2) = -4 + 2 = -2

Getting these?

Answer 33

yes

Answer 34

Then in this question

x2 - 8x + 16

p+q = -8
pq = +16

pq must give a positive number

and p+q = -8

How about -4 * 4 = -16 :( [Not not possible]
But wait, how about -4 * -4 :D

-4 * -4 = 16

-4 + -4 = -8

So p = -4 , q = -4

And then you can put it into the formula

x2 + (p+q)x + pq and factorize it. :)

Answer 35

Ohh okay

Answer 36

Coming to the question in hand,

what are 2 values of 'b' in x2 + bx + 30 so that it is a product of 2 binomials.
Well, they basically want the value of b, for which the polynomial can be factorized.

x2 + bx + 30

b = p+q
30 = pq

So what are the factors of 30?

30 = 6 * 5 {p+q = [b = 11]}
30 = 10 * 3 {p + q = [ b = 13]}
30 = 1 * 30 {p + q =[b = 31]}
30 = 2 * 15 {p + q = [ b= 17]}

Getting this?

Answer 37

yes i am sorry i was taking notes on this so i wont forget.

Answer 38

Okay, just make sure, you get these. And if you did not understand let me help you, I have no problem explaining a hundred times, if you haven't understood. :P

Answer 39

oh no so far i get this, you're teaching this better than my teacher has been teaching us cause her work i dont understand.

Answer 40

Lol, try khanacademy too. They explain it pretty well. :)

Answer 41

okay thanks (:

Answer 42

@mysticghost were is the attached file? Cant find it.