(1) Factor the GCF out of the trinomial on the left side of the equation. (2 points: 1 for the GCF, 1 for the trinomial) 2x²+6x-20=0 (2) Factor the polynomial completely. (4 points: 2 point for each factor) (3) If a product is equal to zero, we know at least one of the factors must be zero. And the constant factor cannot be zero. So set each binomial factor equal to 0 and solve for x, the width of your project. (2 points: 1 point for each factor) (4) What are the dimensions of your project? Remember that the width of your project is represented by x. (2 points: 1 point for each dimension) thank you so much, sorry if it's a little confusing!!

Step-by-step explanation:

(1) Factor the GCF out of the trinomial on the left side of the equation. (2 points: 1 for the GCF, 1 for the trinomial) 2x²+6x-20=0

2x²+6x-20

2(x²+3x-10)

the factors are 2 and (x²+3x-10)

(2) Factor the polynomial completely. (4 points: 2 point for each factor)

2(x²+3x-10)

2(x²-2x+5x-10)

2(x(x-2) + 5(x-2))   group like terms

2(x+5)(x-2)

(3) If a product is equal to zero, we know at least one of the factors must be zero. And the constant factor cannot be zero. So set each binomial factor equal to 0 and solve for x, the width of your project. (2 points: 1 point for each factor)

constant = 2  cannot be zero

the other factors are (x+5) and (x-2)

(x+5)=0 => x= -5

or

(x-2)=0 => x=2

(4) What are the dimensions of your project? Remember that the width of your project is represented by x. (2 points: 1 point for each dimension)

thank you so much, sorry if it's a little confusing!!  

(it is indeed confusing, because physical dimensions cannot be negative)

The dimensions of the project (assumed a rectangle) are +2 and -5

he spend 5.1 minutes chatting with his friends

step-by-step explanation:

2/5+1/2=9/10

6/1-9/10=5 1/10

x= -3 and

step-by-step explanation:

to find the roots of a cubic, use division or factoring to find the factors then set equal to 0. as this is a difference of cubes, it factors into

where a is the cube root of .

here   this is 27 and the cube root of 27 is 3. so we write using the form:

.

set each factor to 0 and solve for x.

x-3=0

x=3

the quadratic expression does not factor and must be solved using the quadratic formula.

where a = 1, b=3, and c=9

this cubic has two complex roots.