$\displaystyle 125x^3-216$

This expression involves the difference of two cubic terms. To factor an expression in this format, we can use a special formula.

$\displaystyle A^3-B^3=(A-B)(A^2+AB+B^2)$

Before we can use this formula, we need to manipulate our original expression to identify $\displaystyle \small A$ and $\displaystyle \small B$.

$\displaystyle 125x^3-216=(5x)^3-6^3$

Comparing this with the formula, $\displaystyle \small A=5x$ and $\displaystyle \small B=6$. Now we can use the formula to factor.

$\displaystyle A^3-B^3=(A-B)(A^2+AB+B^2)$

$\displaystyle (5x)^3-(6)^3=(5x-6)((5x)^2+(5x)(6)+6^2)$

$\displaystyle (5x)^3-(6)^3=(5x-6)(25x^2+30x+36)$

This problem involves the difference of two cubic terms. We need to use a special factoring formula that will allow us to factor this equation.

$\displaystyle a^3-b^3=(a-b)(a^2+ab+b^2)$

But before we can use this formula, we need to manipulate $\displaystyle 343x^3-64y^3$ to make it more similar to the left hand side of the special formula. We do this by making the coefficients (343 and 64) part of the cubic power.

$\displaystyle 343x^3-64y^3=(7x)^3-(4y)^3$

Comparing this with $\displaystyle a^3-b^3$, $\displaystyle a=7x$ and $\displaystyle b=4y$.

Plug these into the formula.

$\displaystyle (7x)^3-(4y)^3=(7x-4y)((7x)^2+(7x)(4y)+(4y)^2)$

$\displaystyle (7x)^3-(4y)^3=(7x-4y)(49x^2+28xy+16y^2)$

Begin by factoring out a 2:

$\displaystyle 2x^{2}+4x-70=2(x^{2}+2x-35)$

Then, we recognize that the trinomial can be factored into two terms, each beginning with $\displaystyle x$:

$\displaystyle 2(x\ \ \ \ )(x\ \ \ \ )$

Since the last term is negative, the signs of the two terms are going to be opposite (i.e. one positive and one negative):

$\displaystyle 2(x+\ \ )(x- \ \ )$

Finally, we need two numbers whose product is negative thirty-five and whose sum is positive two.  The numbers $\displaystyle +7$ and $\displaystyle -5$ fit this description.  So, the factored trinomial is:

$\displaystyle 2(x+7)(x-5)$

You can factor this trinomial by breaking it up into two binomials that lead with $\displaystyle x$:

$\displaystyle (x^2+5x-36) = 0$

$\displaystyle (x \ \ \ \ \ $ (x \ \ \ \ \ \) = 0\)

You will fill in the binomials by finding two factors of 36 that add up to 5. This is achieved with positive 9 and negative 4:

$\displaystyle (x+9)(x-4) = 0$

You can then set each of the two binomials equal to 0 and solve for $\displaystyle x$:

$\displaystyle x+9=0 \ \ \ \ \ \ x-4=0$

$\displaystyle x=-9 \ \ \ \ \ \ x=4$

To factor a polynomial of the form $\displaystyle ax^2+bx+c$, we want to look at the factors of $\displaystyle a$ and the factors of $\displaystyle c$. We want to find the combination of factors which when multiplied and added together give the value of $\displaystyle b$.

In our case, $\displaystyle a=1$, $\displaystyle b=-4$, and $\displaystyle c=-5$.

The factors for $\displaystyle a$ are $\displaystyle (1,1)$.

The factors for $\displaystyle c$ are$\displaystyle (-5,1), (5,-1)$. 

Since $\displaystyle b$ is $\displaystyle -4$ we will want to use the factors $\displaystyle (-5,1)$ because $\displaystyle -5+1=-4$.

Therefore when we put these factors into the binomal form we get,

$\displaystyle (x-5)(x+1)$.

Also see that

$\displaystyle (x-5)(x+1)$ 

will foil out into the original polynomial, as $\displaystyle -5+1=-4$, the coefficient for our $\displaystyle x$ term, and $\displaystyle -5*1=-5$, the constant.

First, factor out the greatest common factor (GCF), which here is $\displaystyle 3x$. If you don't see the whole GCF at once, factor out what you do see (here, either the $\displaystyle 3$ or the $\displaystyle x$), and then check the result to see if any more factors can be pulled out.

Then, to factor a quadratic trinomial, list factors of the quadratic $\displaystyle (x^2)$ term and the constant (no variables) term, then combine them into binomials that when multiplied back out will give the original trinomial.

Here, the quadratic term has only one factorization: $\displaystyle x^2=x \cdot x$.

The constant term has factorizations of $\displaystyle 1 \cdot 24$, $\displaystyle 2 \cdot 12$, $\displaystyle 3 \cdot 8$, and $\displaystyle 4 \cdot 6$.

We know the constant term is positive, so the binomials both have the same operation in them (adding or subtracting), since a positive times a positive OR a negative times a negative will both give a positive result.

But since the middle term $\displaystyle -10x$ is negative, both binomial factors must contain subtraction. And $\displaystyle -4x+(-6x)=-10x$.

$\displaystyle y-7=\frac{4x+6y}{3}$ 

Multiply both sides by 3:

$\displaystyle \frac{3}{1}(y-7)=(\frac{4x+6y}{3})\frac{3}{1}$ 

$\displaystyle 3(y-7)={4x+6y}$ 

Distribute:

$\displaystyle 3y-21={4x+6y}$

Subtract $\displaystyle 3y$ from both sides:

$\displaystyle -21=4x+6y-3y$

Add the $\displaystyle y$ terms together, and subtract $\displaystyle 4x$ from both sides:

$\displaystyle -4x-21=3y$

Divide both sides by $\displaystyle 3$:

$\displaystyle \frac{-4x-21=3y}{3}$

Simplify:

$\displaystyle y=-\frac{4}{3}x-7$

For quadratic equations, you need to factor in order to solve for your variable.  You do this after the equation is set equal to zero.  Thus, you get:

$\displaystyle 5x^2-2x=0$

Next do your factoring:

$\displaystyle 5x^2-2x=0$

into

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

Then, you set each factor equal to $\displaystyle 0$.  Solve each "small" equation:

$\displaystyle x=0$

$\displaystyle 5x-2=0$ or $\displaystyle 5x=2$ or $\displaystyle x=\frac{2}{5}$

BOTH of these are answers to the equation.

For quadratic equations, you need to factor in order to solve for your variable.  You do this after the equation is set equal to zero.  Luckily, this is already done for you!  Thus, start by factoring:

$\displaystyle 3x^2+6x=0$

into

$\displaystyle 3x(x+2)=0$

Then, you set each factor equal to $\displaystyle 0$.  Solve each "small" equation:

$\displaystyle 3x=0$ or $\displaystyle x=0$

$\displaystyle x+2=0$ or $\displaystyle x=-2$

BOTH of these are answers to the equation.

For quadratic equations, you need to factor in order to solve for your variable.  You do this after the equation is set equal to zero.  Thus, you get:

$\displaystyle x^2+7x+6=0$

Next do your factoring.  You know that both groups will be positive.  Also, given that the middle term is $\displaystyle 7$, you only have one possible choice for your factors of $\displaystyle 6$:

$\displaystyle (x+6)(x+1)=0$

Then, you set each factor equal to $\displaystyle 0$.  Solve each "small" equation:

$\displaystyle x+6=0$ or $\displaystyle x=-6$

$\displaystyle x+1=0$ or $\displaystyle x=-1$

BOTH of these are answers to the equation.