From x2 + 6x + 5 = 0 ; To (x + 5) (x + 1) = 0?

Question

How did they get to the second set? I can't remember the formula or rule. I am studying for the gmat and this is the only thing I'm hung up on. I graduated from HS over 10 yrs ago and my brain is dusty! Please help!!!! This is really bugging me that I can't remember this.

Answer 1

Solve "x" in .... x^2 + 6x + 5 = 0

First: multiply the 1st & 3rd coefficient to get "5." Find two numbers that give you "5" when multiplied & "6" (2nd/middle coefficient) when added/subtracted. The numbers are (1 & 5).

Sec: rewrite the equation with the new middle coefficients....

x^2 + x + 5x + 5 = 0

Third: when you have 4 terms --- group "like" terms & factor both sets of parenthesis...

(x^2 + x) + (5x + 5) = 0

x(x+1) + 5(x+1) = 0

(x+5)(x+1) = 0

Fourth: solve the x-variables > set both parenthesis to "0"....

a. x + 5 = 0

*Isolate "x" on one side --- subtract 5 from both sides....

x + 5 - 5 = 0 - 5
x = 0 - 5
x = - 5

b. x + 1 = 0

*Isolate "x" on one side --- subtract "1" from both sides...

x + 1 -1 = 0 -1
x = 0 - 1
x = -1

Solutions: -5 and -1

Answer 2

^ <-- that is the expontent symbol, it basically symnolizes that something is to the power of what ever number ^2 = squared or to the power of 2.

They used the Foil Method (First, Outter, Inner, Last)

x*x= x^2

x*1=x ( you don't need the 1 it is understood automatically)

5*x= 5x

5*1=5

Now you combine like terms

x^2+x+5x+5= x^2 + 6x+5

now you just set it equal to 0 which is the same.

you basically break down the first formula. You don't need the =0 part you can put that in at the end, it will still be the same.

Keep in mind that x times x is always x squared (x ^2)

x^2 + 6x + 5

(x+_) (x+_)

what two numbers multiplied together give u (positive) 5 and added together give you 6? The only option is 5 and 1. Now since they are both positive it doesn't matter where they go. This problem you have is basically factoring trinomials.

Answer 3

checking your old notes huh?

well here's how it's done

that first equation is factorable
it is factorable by using SQT (simple quadratic trinomial)

first think of two numbers that if you multiply them the answer will be the third term(5) and if you add them the answer will be the second term (6)

so the two numbers are 5,1
why?
5*1=5 and,
5+1=6

now place them in each pharenthesis with x in it

(x+5)(x+1)

if you multiply this binomials you will get
x^2+6x+5=0


easy huh?


^_^

Answer 4

You have to distibute the 2nd set into each other.
x*x= x2 (x+5) (x+1)=0
x*1=x
5*x=5x
5*1=5

Now, you have to add like terms (In this problem their are only two)

5x+x=6x

So this gives you

x2+6x+5 Which is the the first set of numbers in the problem

If you still need help email me at [e-mail]

Answer 5

(x+5)(x+1): First you multiply x*x and X*1 so you have x^2+x
then you multiply 5*x and 5*1 and you get 5x+1;

Added together you get x^2+6x+1.

You must multiply each of them out using the distributive property.

Answer 6

A polynomial of grade n is equal to (x-root1)*(x-root2)*....*(x-rootn).

Roots of the cuadratic equation x^2+6x+5=0 are -5 and -1, so you have the expression

(x-(-5)*(x-(-1))=0, or simplifying (x+5)*(x+1) =0

You can easily verify the roots of your equation. Also you can obtain them by solving the equation.

Answer 7

x^2+6x+5=0

x 5
x 1 (cross multiply so u get x times 1 and x times 5)
_________
5x+1x=6x

as you see it adds up to 6x which is in the polynomial x^2+6x+5=0

So u get (x+5)(x+1)

Answer 8

(x + 5) (x + 1)
muliply the first x with both the figures in the second set...
X2 + 1X
then muliply the 5 with both numbers in the second set...
5X + 5
this leaves... x2 + 1x + 5 x + 5...
x2 + 6x + 5

Answer 9

it's a*x^2+b*x+c=0 <=> a(x-x1)(x-x2)=0

x1=[-b+sqrt(b^2-4*a*c)]/2*a
x2=[-b-sqrt(b^2-4*a*c)]/2*a

or x*x+5x+x+5=0
x(x+5)+x+5=0
(x+5)(x+1)=0

Answer 10

thats right


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