Question

Factor completely

12x^2 - 11x + 2

and

x^2 - 49

Answer 1

To factor a trinomial of the form \(ax^2 + bx + c\), you perform the following

1. Find two numbers \(m\) and \(n\) that add to get \(a \times c\) yet also add to get \(b\).

2. Replace \(b\) with \(m + n\)

3. Factor by grouping

To factor a binomial square of the form a^2 - b^2, you simply use the formula
\(a^2 - b^2 = (a + b)(a - b)\)

Answer 2

thanks hero.

Answer 3

So to factor \(12x^2 - 11x + 2\),

1. \(m \times n = 24\)
\(m + n = -11\)

@melf1717, can you think of two numbers that multiply to get 24, yet add to get -11?

Answer 4

for ax^2+bx+c=0 --> (x+x1)(x+x2)=0 when x1 and x2 = \[ = \frac{ -b \pm \sqrt{b ^{2}-4ac} }{ 2a }\]

Answer 5

\[=\frac{ 11\pm \sqrt{121-4(12)(2)} }{ 2(12) }=\frac{ 11\pm5 }{ 24 }\]

Answer 6

x1=16/24 and x2=6/24 then (x1+2/3)(x+1/4)

Answer 7

I get -8 and -3 for the two numbers and then from that we know that
-11 = -(3 + 8)

So the factorization becomes

12x^2 - (3 + 8)x + 2 = 12x^2 - 3x - 8x + 2
= 3x( 4x - 1) - 2(4x - 1)
= (4x - 1)(3x - 2)