2xsquared-11x-21= ?

a=2
b=11
c=21
product= a*c
p=42>21 and 2
2xsquared-21x-2x-21
x(2x-21)-2x-21
this can't be the answer. am i missing a step? please help, thanks

You are not starting out with an equation, since you have no number on the right side, just a ?. Therefore x could be anything

If you are supposed to solve the equation
2x^2 -11 x -21 = 0,
then factor it into
(2x +3)(x-7) = 0
That tells you that the answers are x = -3/2 and x = 7.
If you are trying to use the quadratic formula
x = [-b +/- sqrt(b^2-4ac)]/2a
to get the answers, then b = -11 and c = -21. You got the signs wrong. I have no idea what your a*c product and p are supposed to sigfnify.

possible technique to factor trinomials

in the form of ax^2+bx+c
diamond technique

It looks like you are trying to factorize the given quadratic equation. Let's go through the steps again to ensure we haven't missed anything.

We have the equation: 2x^2 - 11x - 21 = 0.

To factorize the quadratic equation, we need to find two numbers whose sum is equal to the coefficient of the middle term (-11x) and whose product is equal to the product of the coefficient of the quadratic term (2x^2) and the constant term (-21).

The coefficient of the quadratic term is 2, and the constant term is -21. The product of these is 2 * (-21) = -42.

Now, we need to find two numbers whose sum is -11 and whose product is -42. These numbers are -14 and +3.

So, we can rewrite the middle term -11x as -14x + 3x.

Now, let's split the middle term using these values:

2x^2 - 14x + 3x - 21 = 0.

Now, let's factor by grouping:

(2x^2 - 14x) + (3x - 21) = 0.

Factor out the common term:

2x(x - 7) + 3(x - 7) = 0.

Notice that we now have a common factor of (x - 7).

So, we can factor out (x - 7) from both terms:

(x - 7)(2x + 3) = 0.

Now we have factored the equation completely.

The zeros of the equation are x = 7 (from (x - 7) = 0) and x = -3/2 (from (2x + 3) = 0).

Therefore, the factored form of the equation is (x - 7)(2x + 3) = 0.

I hope this helps! Let me know if you have any further questions.