Just started school so im rusty on how to factor these can someone help?

Question

(6p^2)-5p-6=0
The answer is {(-2/3),(3/2)} but i dont know how to get it.
Then this one

(X+1)/3 + (X+2)/3 =5
This is the answer 46/5 (fraction)
And again i dont know how to get! it if someone could please show me I'd appreciate it very much! thanks

(X+2)/3 AHHH i messed up its acutally (X+2)/7 in the problem!!

Answer 1

Quadratics can usually be done in your head with a little practice. (Ax + B) (Cx + D)

For 6p²-5p-6, The minus 6 tells you that the signs of the two terms are opposite, (Ax+B) (Cx-D). The middle term 5 is therefore the difference of ADx - BCx. The 6P² says that AC=6 Likely prospects here are 1-5 and 2-3, so

A few mental experiments lead you to
(3p+2)(2p-3)


(x+1)/3 + (x+2)/3 = 5
multiply each side by 3
(x+1) + (x+2) =15
2x+3 = 15
2x = 12
x = 6 ?

Answer 2

6p^2-5p-6=0 changes to
(3p+2)*(2p-3)=0, you just have to guess and check on the factoring. Then you set up two different problems: 3p=-2, p=-2/3; and 2p=3, p=3/2.

for the second problem, multiply both sides by 3 to get:
x+1+x+2=15 --> 2x+3=15 --> 2x=12, x=6. the answer should not be a fraction. If x=6, then (6+1)/3 + (6+2)/3=5 --> 7/3 + 8/3 =5 --> 15/3=5.

Answer 3

a =6, b = -5 c = -6

ac = -36

think of two numbers whose product is -36 and sum is -5(i.e.,b)
by trial and error the numbers are (-9 ) and 4
use this to split the p term

(6p^2)-9p +4p -6=0
3p(2p - 3) +2(2p - 3) =0

(2p - 3)(3p+2) = 0

p = (3/2) , (-2/3)



(2)
L.C.M is 21,

(X+1)/3 + (X+2)/7 =5

[ { 7*(X+1) +3*(X+2) }/21 ] = 5

[ { 10x + 13} / 21 ] = 5

10x+13 = 105

10x = 105 - 13

10x = 92

x = 92/10

or x = 46/5 (cancelling off 2)

Answer 4

This is a quadratic function. The above solution equals (3p+2)(2p-3)=0. Now set each part of the inequality equality to zero, So 3p+2=0 --> p = -2/3. For the second part, p = 3/2.

Answer 5

For quadratic equation, there is a formular to solve the coefficient.


Ax^2 + bx + c

x =

(- b + squareroot( b^2 - 4ac ) ) / 2a and

(- b - squareroot( b^2 - 4ac ) ) / 2a