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Mathematics 9 Online
OpenStudy (andrewkaiser333):

What is the factored form of x^2 + 6x + 8? (x + 5)(x + 3) (x + 4)(x + 2) I think it is this one atm can you check (x + 7)(x + 1) (x + 3)(x + 3)

OpenStudy (anonymous):

is this a homework problem?? somebody asked the same question

OpenStudy (andrewkaiser333):

yes it is but i am wanting to see if i got it right

OpenStudy (anonymous):

yes you got it right

OpenStudy (andrewkaiser333):

thanks

OpenStudy (andrewkaiser333):

what about this one What is the factored form of x2 – 7x + 12? (1 point) (x – 5)(x – 3) I think this is it (x – 6)(x – 1) (x – 5)(x – 2) (x – 4)(x – 3)

OpenStudy (anonymous):

(x-5)(x-3) = x(x-3) -5(x-3) = x^2 -3x -5x +15 = x^2-8x +15 how can this be equal to x^2-7x+12?

OpenStudy (anonymous):

i gave you the principle. try to figure out the pattern in it for every the equation of form(x+a)(x+b)

OpenStudy (andrewkaiser333):

sorry have not done this before so i am trying to figure it out

OpenStudy (anonymous):

remeber that x is real number even though it looks very complicated.. so it has all the property of real number and that means you can do the same operations for it as you do to the real numbers

OpenStudy (anonymous):

expand (x+a)(x+b) and you will see what i mean

OpenStudy (andrewkaiser333):

can you give me a hint so i can try to get it

OpenStudy (anonymous):

well do you see this? (2+3)(4+6)=(2+3)*4 + (2+3)*6 =2*4+3*4 +2*6+3*6

OpenStudy (andrewkaiser333):

i am confused

OpenStudy (anonymous):

well do you see this? 2*(3+4)=2*3+2*4

OpenStudy (andrewkaiser333):

what you mean

OpenStudy (anonymous):

let's just use this equation.. 5(4+6)=5*4+5*6

OpenStudy (anonymous):

i am asking that you know that operation that the equation above holds

OpenStudy (andrewkaiser333):

i said i don't understand

OpenStudy (anonymous):

if you don't understand this 5(4+6)=5*4+5*6, you don't understand your problem

OpenStudy (andrewkaiser333):

i know that is why i asked for you to help

OpenStudy (anonymous):

well ok. memorize this a(b+c)=ab+ac

OpenStudy (andrewkaiser333):

ok

OpenStudy (anonymous):

it means when you add b &c and multiply it by a, you get the same results as a times b plus a times c

OpenStudy (anonymous):

a, b, c above are real numbers.. like 1, 2, 3, 4, 1/2, 1/3 .... stuffs

OpenStudy (andrewkaiser333):

ok

OpenStudy (anonymous):

so let's accept that this rule works and comeback to our problem (x+a)(x+b) first of all (x+a) is real number right? so by our rule, (x+a)(x+b)=(x+a)x+(x+a)b again, (x+a)x+(x+a)b=xx+ax+xb+ab =x^2+ax+bx+ab =x^2+(a+b)x+ab

OpenStudy (anonymous):

i only used the a(b+c)=ab+ac rule.. try to follow it.

OpenStudy (andrewkaiser333):

ok

OpenStudy (anonymous):

so it says, (x+a)(x+b) equals x^2+(a+b)x+ab

OpenStudy (andrewkaiser333):

would it be C then??

OpenStudy (anonymous):

no no i only used a(b+c)=ab+ac to show you the pattern

OpenStudy (anonymous):

the a, b used in (x+a)(x+b) are not same as the a(b+c) ...

OpenStudy (anonymous):

it just means random number

OpenStudy (anonymous):

every number

OpenStudy (andrewkaiser333):

ugh i am confused

OpenStudy (anonymous):

ok just for get about every thing and remember a(b+c)=ab+ac

OpenStudy (anonymous):

insert a=(x+p) , b=x , c=q in that a(b+c)=ab+ac would you do that?

OpenStudy (anonymous):

what does it tell you

OpenStudy (andrewkaiser333):

i can't think straight my mind is messed up now i think i fixed to many computers that i can't think right

OpenStudy (anonymous):

i am just asking you to plug (x+p), x, q into the a,b,c position is the equation a(b+c)=ab+ac this is just matching the same picture business

OpenStudy (anonymous):

just plug it in!!!

OpenStudy (andrewkaiser333):

i can't i am not able to think right

OpenStudy (anonymous):

ok rest for while. the answer is (x+p)(x+q)=(x+p)x+qx anyways

OpenStudy (andrewkaiser333):

i am going my head hurts

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