  ## From a handpicked tutor in live 1-to-1 classes Basics of Algebra

## 1. Basic Factoring

$\text{ Solve for } x:$ $x^2+6x+8 = 0$

To solve this quadratic equation by factoring, we can look for two numbers that multiply to give 8 and add up to 6. These numbers are 2 and 4.

Therefore, we can rewrite the equation as

$(x + 2)(x + 4) = 0$
Setting each factor equal to zero, we have
$x + 2 = 0 \text{ and } x + 4 = 0$ $\Rightarrow x = -2 \text{ and } x = - 4$

Hence, the solutions to the equation $$x^2 + 6x + 8 = 0$$ are $$x = -2$$ and $$x = -4$$.

## 2. Basic Factoring with negative numbers

$\text{ Solve for } x:$ $x^2 -6x+8 = 0$

To solve this quadratic equation by factoring, we are still looking for two numbers that multiply to give 8 but this time they add up to -6. These numbers are -2 and -4.

Therefore, we can rewrite the equation as

$(x - 2)(x - 4) = 0$
Setting each factor equal to zero, we have
$x - 2 = 0 \text{ and } x - 4 = 0$ $\Rightarrow x = 2 \text{ and } x = 4$

Hence, the solutions to the equation $$x^2 - 6x + 8 = 0$$ are $$x = 2$$ and $$x = 4$$.

## 2. Basic Factoring with a constant (Part A)

$\text{ Solve for } x:$ $3(x^2 -6x+8) = 0$

We ignore the 3 and start by factoring the quadratic expression $$x^2 - 6x + 8$$. As discussed above, the factored form is $$(x - 2)(x - 4)$$.

So, we have
$3(x^2 -6x+8) = 3(x - 2)(x - 4)$
Setting each factor equal to zero, we have
$x - 2 = 0 \text{ and } x - 4 = 0$ $\Rightarrow x = 2 \text{ and } x = 4$

Hence, the solutions to the equation $$3(x^2 - 6x + 8) = 0$$ are $$x = 2$$ and $$x = 4$$.

## 2. Basic Factoring with a constant (Part B)

$\text{ Solve for } x:$ $4x^2 - 24x + 32 = 0$
Anytime, we see a number in front of the $$x^2$$ term, we want to see if we can factor it out. In this example, we can first factor out the 4 which makes the equation much easier to solve.
$4x^2 - 24x + 32 = 4(x^2 -6x+8)$

Now as in the previous problem, we ignore the 4 and start by factoring the quadratic expression $$x^2 - 6x + 8$$. As discussed above, the factored form is $$(x - 2)(x - 4)$$.

So, we have
$4(x^2 -6x+8) = 4(x - 2)(x - 4)$
Setting each factor equal to zero, we have
$x - 2 = 0 \text{ and } x - 4 = 0$ $\Rightarrow x = 2 \text{ and } x = 4$

Hence, the solutions to the equation $$4x^2 - 24x + 32 = 0$$ are $$x = 2$$ and $$x = 4$$.

## 2. Factoring out the gcf

$\text{ Solve for } x:$ $4x^3 - 24x^2 + 32x = 0$
In this advanced version of the previous problem, we aim to factor out common numbers and also variables. By observing the terms, we first factor out the 4 and also the $$x$$ which makes the equation much easier to solve.
$4x^3 - 24x^2 + 32x = 4x(x^2 -6x+8)$

Now as in the previous problems, we ignore the $$4x$$ term and start by factoring the quadratic expression $$x^2 - 6x + 8$$. As discussed above, the factored form is $$(x - 2)(x - 4)$$.

So, we have
$4x(x^2 -6x+8) = 4(x - 2)(x - 4)$
Setting each factor equal to zero, we have
$x - 2 = 0 \text{ and } x - 4 = 0$ $\text{ also } x = 0$ $\Rightarrow x = 0, x = 2 \text{ and } x = 4$

Hence, the solutions to the equation $$4x^3 - 24x^2 + 32x = 0$$ are $$x = 0, x = 2$$ and $$x = 4$$.

## 2. Two minor modifcations to the quadratic

$\text{ Solve for } x:$ $-x^2 + 6x - 8 = 0$
Any time we see a negative sign in front of the $$x^2$$, we can simplify it by multiplying every term by -1. Now the$$x^2$$ term becomes positive. Then, we can proceed to solve the quadratic equation as usual.
$-x^2 + 6x - 8 \rightarrow x^2 - 6x + 8$

Now as discussed above, we factor the quadratic as $$(x - 2)(x - 4)$$.

Setting each factor equal to zero, we have
$x - 2 = 0 \text{ and } x - 4 = 0$ $\Rightarrow x = 2 \text{ and } x = 4$

Hence, the solutions to the equation $$-x^2 + 6x - 8 = 0$$ are also $$x = 2$$ and $$x = 4$$.

$\text{ Solve for } x:$ $x^2 - 6x = - 8$
To solve a quadratic by factoring, we bring all terms to one side and get a 0 on the right hand side. So we add 8 to both sides.
$x^2 - 6x = - 8$ $\Rightarrow x^2 - 6x + 8 = 0$

Then, we can proceed to solve the quadratic equation as usual. So, as before, we factor the quadratic as $$(x - 2)(x - 4)$$.

Setting each factor equal to zero, we have
$x - 2 = 0 \text{ and } x - 4 = 0$ $\Rightarrow x = 2 \text{ and } x = 4$

Hence, the solutions to the equation $$-x^2 + 6x - 8 = 0$$ are also $$x = 2$$ and $$x = 4$$.

## 2. Missing Constant Term

$\text{ Solve for } x:$ $x^2 - 6x = 0$
When the constant term is missing, we simply factor out the $$x$$ term
$x^2 - 6x = 0 \rightarrow x(x - 6) = 0$

Setting each factor equal to zero, we have:

$x = 0 \text{ or } x - 6 = 0$

Therefore, the solutions to the equation $$x^2 - 6x = 0$$ are $$x = 0$$ and $$x = 6$$.

## 2. Missing $$x$$ Term

$\text{ Solve for } x:$ $x^2 - 16 = 0$
When the $$x$$ term is missing, we can start by adding 16 to both sides of the equation.
$x^2 - 16 + 16 = 0 + 16$ $\Rightarrow x^2 = 16$
Now, we can take the square root of both sides to find the solutions.
$x = \pm 4$
Therefore, the solutions to the equation $$x^2 - 16 = 0$$ are $$x = 4$$ and $$x = -4.$$

## 2. The $$ax^2 + bx + c$$ method

All quadratics can be written as $$ax^2 + bx + c = 0$$.

When we compare equation $$3x^2 + 5 = 6x$$ with $$ax^2 + bx + c = 0$$, we get
$3x^2 + 6x + 5 = 0$ $\Rightarrow a = 3, b = 6, c = 5$

In the equation $$3x^2 + 5 = 6x$$, we first move the $$6x$$ to the left to get the equation $$3x^2 -6x + 5 = 0$$. Then we get

$3x^2 + 5 = 6x$ $\Rightarrow 3x^2 -6x + 5 = 0$ $\Rightarrow a = 3, b =-6, c = 5$

$x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$
$2x^2 - 5x + 3 = 0$ $\Rightarrow a = 2, b = -5, c = 3$
Substituting the values $$a = 2, b = -5, c = 3$$ into the formula, we get:
$x = \frac{{-(-5) \pm \sqrt{{(-5)^2 - 4(2)(3)}}}}{{2(2)}}$
$x = \frac{{5 \pm \sqrt{{25 - 24}}}}{{4}}$ $x = \frac{{5 \pm \sqrt{1}}}{{4}}$ $x = \frac{{5 \pm 1}}{{4}}$ $\Rightarrow x = \frac{3}{2} \text{ and } x = 1$
Therefore, the solutions to the equation are $$x = \frac{3}{2}$$ and $$x = 1$$ 6 more sections Finding the Vertex: Standard Form y = x^2 - 6x + 8 Finding the Vertex: Intercept Form y = (x-2)(x-4) Finding the Vertex: Vertex Form y = (x-3)^2 + 2 How many real solutions does this equation have? x^2 -6x + 8 = 0 How many real solutions does this equation have: x^2 -6x + 9 = 0 How many real solutions does this equation have: x^2 -6x + 10 = 0