Algebra 1 help factoring trinomials ?!?

The first one has a common factor of 3, which simplifies it to

3(x^2 + x - 20) = 0
3(x + 5)(x - 4) = 0

The 5 & -4 appear here because we are looking for two numbers which multiply to -20 (constant tern, without x in it) and add to +1 (coefficient of x). The only two numbers are +5 & -4.
It is an equation (= 0) so you can solve with x = -5 or x = 4.

The second one is harder & a trick is required. For this sort of trinomial (with a coefficient of x^2 which is bigger than 1 and it won't factor out as in Q1).

The trick is multiply coeff of x^2 (The 2) by constant (the -54) to get -108. You now need to find two numbers which multiply to this -108 but add to the middle number (the coeff of x "3").
These turn out to be +12 and -9.

We now re-write the middle term as +12x - 9x (as this is still equal to 3x). So we get

2x^2 + 12x - 9x - 54 = 0 (I know it is now 4 terms but we can find the common factor for the first two & separately the second two)

2x(x + 6) - 9(x + 6) = 0

You will always get the same bracket appearing here, which itself a common factor

(x + 6)(2x - 9) = 0

The second bracket are just the numbers in front of the repeated bracket.

Solving

x = - 6 or x = 9/2

This method requires practice but is a neat trick which always works with this type of quadratic.

Good luck!

for type one: you could pull out the straight forward ingredient of 2x^2 to get: 2x^2 (x^2-8x+15) To ingredient what's in paranthesis, we would desire to first think of: What 2 numbers elevated jointly make 15 and added jointly make -8? After somewhat of concept, we see that -5*-3 is 15 and -5 plus -3 is -8, so our 2 aspects are: (x-5)(x-3) SO the appropriate factored answer is: 2x^2(x-5)(x-3) :) for the 2d difficulty, we are in a position to pull out a straight forward ingredient of 7x to get: 7x(x^2 +7x+10) Now enable us to think of what 2 numbers elevated jointly make 10 and added jointly make 7? We arise with 5*2=10 and 5+2 = 7, so our 2 aspects are: (x+5)(x+2) So the appropriate answer could be: 7x(x+5)(x+2) :)

1)
3x^2 + 3x - 60 = 0
3x^2 + 15x - 12x = 0
(3x + 15)(x - 4) = 0

3x + 15 = 0
3x = -15
x = -15/3
x = -5

x - 4 = 0
x = 4

∴ x = -5 , 4

- - - - - - - - - - - - - - -

2)
2x^2 + 3x - 54 = 0
2x^2 - 9x + 12x - 54 = 0
(2x - 9)(x + 6) = 0

2x - 9 = 0
2x = 9
x = 9/2
x = 4 1/2 (4.5)

x + 6 = 0
x = -6

∴ x = 4 1/2 (4.5) , -6

3x^2 + 3x - 60 = 0

Firstly, since you have a number in front of the 3x you have to multiply that by 60. But wait, since all three of them can be reduced, lets divide them all by 3.
So your new equation is:
3(x^2 + x - 20)= 0
[keep in mind since you divided them all by three, you have to keep the three there, outside of the equation. & it will stay! it's still part of the equation.]

Now, you factor out. You need to find factors that multiply to -20 and add to 1. Which would be a -4 and a 5. (if you were to add -4 + 5 = 1. and -4 x 5 = -20.

So your factors are -4 & 5. Now you need to fill them in into an equation. Back to the equation: 3(x^2 + x - 20)= 0
since you have x squared, you'll need to part those into each of your parenthesis. And add on your factors. So your final answer would be:
3(x-4)(x+5)

3x^2+3x-60=0
dividing by 3
x^2+x-20=0
x^2+5x-4x-20=0
(x-5)(x-4)=0