CPhill

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 #1
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First one....a coefficient can only be  "attached" to a variable....."10"  has no variable attached to it...so...it is not a coefficient

 

Second one

 

      -x^2  - 5x   -    6

 

+ (  2x^2  - 7x   +  8 )

________________

     x^2    - 12x  +  2

 

 

Third one  -  9x^2 - 60x + 100

This is a perfect square trinomial....take the square root of the first term  = 3x

Take the square root of the last term  = 10

Since we have a  " - "   on the second term....that's the sign we use in factoring....so we have

(3x^2  - 10) (3x^2 - 10 )  = (3x - 10 )^2

 

 

Fourth one -

 

(x - 3)^2  + 8  = 12     subtract 8 from both sides

(x - 3)^2   = 4             take both square roots on the right side

x - 3  = ±√4

x - 3  =  ±2          add 3 to both sides

x  = ±2  + 3.....so we have

x  = 2 + 3  = 5      or

x  = -2 + 3   = 1

So...the two solutions are   x = 1  and x  = 5

 

 

Last one

 

x^2 + 14x  =- 48     add 48 to both sides

x^2 + 14x  + 48  = 0

We are looking for two  numbers to factor with that multiply to 48  and sum to -14....note....-8  and -6  fit the bill  !!!

So....factor this as

(x - 8) (x - 6)  =  0

Set each factor to 0 and solve for x  and we have

 

x  - 8  =  0                              x  - 6  = 0

add 8 to both sides              add 6 to both sides

 

x  = 8                                        x  = 6                  are the two solutions

 

 

 

cool cool cool

17.04.2018