$\displaystyle 7x^2+56+C$

Factor out the 7:

$\displaystyle 7\left (x^2+8x+\frac{C}{7} \right )$

Take the 8 from the x-term, cut it in half to get 4, then square it to get 16.  Make this 16 equal to C/7:

$\displaystyle 16=\frac{C}{7}$

Solve for C:

$\displaystyle C=112$

We can factor this trinomial using the FOIL method backwards. This method allows us to immediately infer that our answer will be two binomials, one of which begins with $\displaystyle 3x$ and the other of which begins with $\displaystyle x$. This is the only way the binomials will multiply to give us $\displaystyle 3x^{2}$.

The next part, however, is slightly more difficult. The last part of the trinomial is $\displaystyle -2$, which could only happen through the multiplication of 1 and 2; since the 2 is negative, the binomials must also have opposite signs.

Finally, we look at the trinomial's middle term. For the final product to be $\displaystyle -x$, the 1 must be multiplied with the $\displaystyle 3x$ and be negative, and the 2 must be multiplied with the $\displaystyle x$ and be positive. This would give us $\displaystyle \left ( -3x\right )+\left ( 2x\right )$, or the $\displaystyle -x$ that we are looking for.

In other words, our answer must be 

$\displaystyle \left ( 3x+2\right )\left ( x-1\right )$ 

to properly multiply out to the trinomial given in this question.

$\displaystyle 2x^{2}+15x+25=0$

1) The first step would be to simplify, but since 2, 15, and 25 have no common factors greater than 1, simplification is impossible.

Now we factor. Multiply the first coefficient by the final term and list off the factors.

2 * 25 = 50

Factors of 50 include:

1 + 50 = 51

2 + 25 = 27

5 + 10 = 15

2) Split up the middle term to make factoring by grouping possible.

$\displaystyle 2x^{2}+10x+5x+25=0$

Note that the "2" and the "10," and the "5" and the "25," have to go together for factoring to come out with integers. Always make sure the groups actually have a common factor to pull.

3) Pull out the common factors from both groups, "2x" from the first and "5" from the second.

$\displaystyle 2x(x+5)+5(x+5)=0$

4) Factor out the "(x+5)" from both terms.

$\displaystyle (2x+5)(x+5)=0$

5) Set each parenthetical expression equal to zero and solve.

2x + 5 = 0,  x = –5/2

x + 5 = 0, x = –5

$\displaystyle x^2 + 0x -9$

$\displaystyle -3 \times3=9$

$\displaystyle -3+3=0$

$\displaystyle (x-3)(x+3) = x^2 – 9$

When we attempt to factor a quadratic, we must first look for the factored numbers. When quadratics are expressed as $\displaystyle ax^2+bx+c$ the factored numbers are $\displaystyle a$ and $\displaystyle c$. Since $\displaystyle a=1$, we know the factors for 1 are 1 and 1. So we know the terms will be $\displaystyle \left ( 1x+ \_\_\_ \right )(1x+\_\_\_)$

Looking at our constant, $\displaystyle c$, we see a positive 6. So 6 factors into either 2 and 3  or 1 and 6 (since $\displaystyle 2\cdot 3=6$ and $\displaystyle 1\cdot 6=6$). Since our constant is a positive number, we know that our factors are either both positive, or both negative. (Note: you should know that 2 negative numbers multiplied becomes a positive number).

So to figure out what we must use we look at the $\displaystyle b$ part of the quadratic. We are looking for 2 numbers which add up to our $\displaystyle b$. So, 1 and 6 do not work, since $\displaystyle 1+6=7$. But, 2 and 3 are perfect since $\displaystyle 2+3=5$.

But, since our $\displaystyle b$ is a negative 5, we know we must use negative numbers in our factored expression. Thus, our factoring must become

 $\displaystyle \left ( 1x+-2 \right )(1x+-3)$

or

$\displaystyle \left ( x-2 \right )\left ( x-3 \right )$

Using the FOIL rule, only $\displaystyle (3x+4)(x-2)$ yields the same polynomial as given in the question.

When asked to factor a difference of squares, the solution will always be the square roots of the coefficients with opposite signs in each pair of parentheses.

First, factor the numerator: $\displaystyle (x+5)(x-3)$.  

Now your expression looks like 

$\displaystyle \frac{(x+5)(x-3)}{x+5}$

Second, cancel the "like" terms - $\displaystyle (x+5)$ - which leaves us with $\displaystyle x-3=10$.  

Third, solve for $\displaystyle x$, which leaves you with $\displaystyle x=13$. 

Because the $\displaystyle x^2$ term has a coefficient, you begin by multiplying the $\displaystyle a$ and the $\displaystyle c$ terms ($\displaystyle ax^2+bx+c$) together: $\displaystyle 2\times-3=-6$. 

Find the factors of $\displaystyle -6$ that when added together equal the second coefficient (the $\displaystyle b$ term) of the polynomial: $\displaystyle -5$. 

There are four factors of $\displaystyle 6$: $\displaystyle 1, 2, 3, 6$, and only two of those factors, $\displaystyle 1, 6$, can be manipulated to equal $\displaystyle -5$ when added together and manipulated to equal $\displaystyle -6$ when multiplied together: $\displaystyle 1, -6$ 

$\displaystyle 1+-6=-5, 1\times-6=-6$

Step 1: Subtract 30 from both sides of the equation in order to make the equation equal to 0.

$\displaystyle 3x^{2} - 9x = 30 \rightarrow 3x^{2} - 9x - 30 = 0$

Step 2: Factor out a 3.

$\displaystyle \rightarrow 3(x^{2} - 3x - 10) = 0$

Step 3: Factor the trinomial.

$\displaystyle \rightarrow 3(x + 2)(x - 5) = 0$

At this step you set both factors equal to 0 and solve for $\displaystyle x$.

$\displaystyle x + 2 = 0 \ri \rightarrow x = -2$

$\displaystyle x - 5 = 0 \rightarrow x = 5$