Quick way to factor this polynomial?

Sure the key is the b term +5. Your job is to separate the factors of 33 in such a way that 2 times one factor yields a difference of 5 whan the "i" & "o" terms are added together. Sooo, 3+11 seem to do the trick

(2x+11)(x-3)=0

2x² + 5x - 33 = 0
2x² + 5x = 33
x² + 5/2x = 33/2
x² + 5/4x = 33/2 + (5/4)²
x² + 5/4x = 264/16 + 25/16
(x + 5/4)² = 289/16
x + 5/4 = ± 17/4

Factors starting from the last line above:
= x + 5/4 - 17/4, = x - 12/4, = x - 3
= x + 5/4 + 17/4, = x + 22/4, = x + 11/2, = 2x + 11

Answer: (x - 3)(2x + 11)—there's no shorter way I could think of.

Proof:
= (x - 3)(2x + 11)
= x(2x) + x(11) - 3(2x) - 3(11)
= 2x² + 11x - 6x - 33
= 2x² + 5x - 33

for the primary one i could distribute the whole lot out and get n^two+sixteen n+sixty three which elements to (n+7)(n+nine) for the moment i could aspect through grouping (y^three-y^two) + (3y-three) and aspect each and every of the ones y^two(y-a million)+three(y-a million) and those mix to (y^two-three)(y-a million)

2x^2+11x -6x-33=0 :x[2x+11]-3[2x+11]=0 ;[x-3][2x+11]=0 x-3=0 and2x+11=0 x=3 and x=-11/2

Quadratic formula / completing the square

2x^2 + 5x - 33 = 0

What I do is:
- Make a big 'X'
- On the top, write '-66' [co-efficient of 'x^2' multiplied by the last number]
- On the bottom, write '5' [co-efficient of 'x']
- Now... Find two numbers that multiply to -66 and add to 5

Numbers: 11 and -6

Now re-write the question as:

2x^2 -6x + 11x - 33 = 0 [Take '2x' common; Take '11' common]
2x(x - 3) + 11(x - 3) = 0 [Take '(x - 3)' common]
(x - 3) (2x + 11) = 0

Now... Solve for 'x'

Make two different equations --

(x - 3) = 0 and (2x + 11) = 0

First equation --

(x - 3) = 0
x - 3 = 0
x = 3 [One answer]

Second equation --

(2x + 11) = 0
2x + 11 = 0
2x = -11
x = -11/2 [Second answer]

Answers:
x = 3
x = -11/2

use the quad formula
x = (-5 +/- sqrt(25 + 264))/4 = (-5+/- 17)/4 = -11/2 or 3

listing the factor of 66 is just as easy in this case, -6 +11
(2x +11)(x - 3) = 0
x = -11/2 or 3

(2x + 11)(x – 3)

There are only two combinations: 1 and 33 or 3 and 11. Only one way to get a 2 (1)(2)

You start with (2x + __)( x – __)

Now use logic. Since 5 is small, you need two factrs of 33 that are fairly close, namely 3 and 11 (closer that 1 and 33). Done.