+129852 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 )
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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
