A(x) = 2 (3x - 2)(x - 5)

Quadratic polynomial can be factored using the transformation ax  

2

+bx+c=a(x−x  

1

)(x−x  

2

), where x  

1

 and x  

2

 are the solutions of the quadratic equation ax  

2

+bx+c=0.

−x  

2

−3x+5=0

All equations of the form ax  

2

+bx+c=0 can be solved using the quadratic formula:  

2a

−b±  

b  

2

−4ac

. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

x=  

2(−1)

−(−3)±  

(−3)  

2

−4(−1)×5

Square −3.

x=  

2(−1)

−(−3)±  

9−4(−1)×5

Multiply −4 times −1.

x=  

2(−1)

−(−3)±  

9+4×5

Multiply 4 times 5.

x=  

2(−1)

−(−3)±  

9+20

Add 9 to 20.

x=  

2(−1)

−(−3)±  

29

The opposite of −3 is 3.

x=  

2(−1)

3±  

29

Multiply 2 times −1.

x=  

−2

3±  

29

Now solve the equation x=  

−2

3±  

29

 when ± is plus. Add 3 to  

29

.

x=  

−2

29

+3

Divide 3+  

29

 by −2.

x=  

2

−  

29

−3

Now solve the equation x=  

−2

3±  

29

 when ± is minus. Subtract  

29

 from 3.

x=  

−2

3−  

29

Divide 3−  

29

 by −2.

x=  

2

29

−3

Factor the original expression using ax  

2

+bx+c=a(x−x  

1

)(x−x  

2

). Substitute  

2

−3−  

29

 for x  

1

 and  

2

−3+  

29

 for x  

2

.

−x  

2

−3x+5=−(x−  

2

−  

29

−3

)(x−  

2

29

−3

)

EVALUATE

5−3x−x  

2Quadratic polynomial can be factored using the transformation ax  

2

+bx+c=a(x−x  

1

)(x−x  

2

), where x  

1

 and x  

2

 are the solutions of the quadratic equation ax  

2

+bx+c=0.

−x  

2

−3x+5=0

All equations of the form ax  

2

+bx+c=0 can be solved using the quadratic formula:  

2a

−b±  

b  

2

−4ac

. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

x=  

2(−1)

−(−3)±  

(−3)  

2

−4(−1)×5

Square −3.

x=  

2(−1)

−(−3)±  

9−4(−1)×5

Multiply −4 times −1.

x=  

2(−1)

−(−3)±  

9+4×5

Multiply 4 times 5.

x=  

2(−1)

−(−3)±  

9+20

Add 9 to 20.

x=  

2(−1)

−(−3)±  

29

The opposite of −3 is 3.

x=  

2(−1)

3±  

29

Multiply 2 times −1.

x=  

−2

3±  

29

Now solve the equation x=  

−2

3±  

29

 when ± is plus. Add 3 to  

29

.

x=  

−2

29

+3

Divide 3+  

29

 by −2.

x=  

2

−  

29

−3

Now solve the equation x=  

−2

3±  

29

 when ± is minus. Subtract  

29

 from 3.

x=  

−2

3−  

29

Divide 3−  

29

 by −2.

x=  

2

29

−3

Factor the original expression using ax  

2

+bx+c=a(x−x  

1

)(x−x  

2

). Substitute  

2

−3−  

29

 for x  

1

 and  

2

−3+  

29

 for x  

2

.

−x  

2

−3x+5=−(x−  

2

−  

29

−3

)(x−  

2

29

−3

)

EVALUATE

5−3x−x  

2

Step-by-step explanation:

hey there!

if she's saving for six months, let's divide the total number we need for the trip by six, dividing it into the amount that needs to be saved for each month.

500 ÷ 6 =

she needs to save at least this amount each month, so we should use this symbol: ≥

let x be the amount she saves each month.

your final answer is x ≥

hope this !

step-by-step explanation: