Factoring quadratic expressions with leading coefficient 1
Factor the following expressions
1. a^2+6a+8
2. b^2+3b+2
3. c^2-11c+24
4. d^2-9d+18
5. e^2-12e+27
6. f^2+8f+15
7. g^2-20g+36

Factoring Trinomials: ax2 + bx + c
Factor
1.〖2a〗^2+15a+25
2. 3b^2+13b+12
3. 〖4c〗^2-9c-9
4.〖4d〗^2-3d-10
5. 3e^2-37e+12
6. 〖2f〗^2+9f+10
7 .〖4g〗^2-17g+15

8.〖5c〗^2+13c+6
9. 2b^2-5b+2
10 3c^2-2c-1

step-by-step explanation:

30% of $420 is 0.30($420) = $126 and $420 + $126 = $546.

or, 130% of $420 is 1.3($420) = $546.

notice that this is not the $600 original list price with which we started.

why doesn't this work?

the method does not work because 30% of $420 is not the same amount as the 30% of $600. it may seem obvious when stated this way, but it is a very common error made on problems of this type. we cannot just use the same percent and get back to the original amount because we are taking that percent of a different value.

to work problems of this type correctly, it is generally best to write an equation for the situation and then solve.

example: since (100% – 30%)(original price) = (sale price)

70% of the   (original price) is $420.

0.7x   = 420

equation image indicator

x = 600

the original cost was $600.

we may also solve this problem using the two-step method. working the problem in two steps requires us to use more algebra skills. we have to know that x = 1x (identity property of multiplicationtext annotation indicator) and that 1x – 0.3x = 0.7x (subtracting like terms). then we solve the problem as follows:

(original price) – 30%(original price) = sale price

x – 0.3x = 420

0.7x = $420

equation image indicator

x = 600

the original cost was $600.  

note that with both of the above methods we do obtain the correct original amount.

example: after a 5% pay raise, hermione is earning $22,680 per year. what was she earning before the pay raise?

105% of (original salary) = current salary

1.05x = 22,680

equation image indicator

x = 21,600

hermione earned $21,600 per year before the pay raise.

this is an example so maybe you will get how to at least do it!

the rocket cannot penetrate the ground, in a contextual sense, therefore the roots of this equation should be the domain of this function. i'll let you find the roots of this quadratic equation (hint: use the quadratic formula:  

1- c

2-a

step-by-step explanation:

mark brainliest!

-4, -2, 1, 4

step-by-step explanation:

domain = x values

range = y values

m=-16

step-by-step explanation: