\(\displaystyle q^2+9q+14=0\)\(\displaystyle Factored:(q+2)(q+7) = 0\)\(\displaystyle q+2 = 0 \;or\; q+7 = 0\)

\(\displaystyle First\;equation:q+2 = 0\)

\(\displaystyle q=-2\)

\(\displaystyle Second \;equation: q+7=0\)
\(\displaystyle q=-7\)

\(\displaystyle q=-7,-2\)

\(\displaystyle Factor \;t^2 + 6t + 8 = 0\)

\(\displaystyle (t+4)(t+2)=0\)

\(\displaystyle t=-4,-2\)

\(\displaystyle x^2 + 7x + 15 = 5\)

\(\displaystyle x^2 + 7x + 10 = 0\)

\(\displaystyle (x+5)(x+2)=0\)
\(\displaystyle x=-2,-5\)

\(\displaystyle \small x^2+6x+8=0\)

Solve by factoring. We need to find two factors that multiply to eight and add to six.

\(\displaystyle \small 2*4=8\ \text{and}\ 2+4=6\)

\(\displaystyle \small x^2+6x+8=(x+2)(x+4)=0\)

One of these factors must equal zero in order for the equation to be true.

\(\displaystyle \small x+2=0\ or\ x+4=0\)

\(\displaystyle \small x=-2\ or\ x=-4\)

First, we must factor out any common factors between the two terms.  Both 3 and 12 share a factor of 3, so we can "take" 3 out, like this:

\(\displaystyle 3(x^{2}-4)\).

Inside the parentheses, it becomes clear that this is a difference of squares problem (a special factor), which can be solved with the equation

\(\displaystyle a^{2}-b^{2}=(a-b)(a+b)\).

Thus, \(\displaystyle x^{2}-2^{2}=(x-2)(x+2)\). 

Now, we can set each factor to 0, and solve for \(\displaystyle x\):

\(\displaystyle x-2=0\)

\(\displaystyle x=2\)

and

\(\displaystyle x+2=0\)

\(\displaystyle x=-2\).

We'll call one of the integers \(\displaystyle x\), which makes the other integer \(\displaystyle x+4\). The product between \(\displaystyle x\) and \(\displaystyle x+4\) is \(\displaystyle -4\).  This is written as

\(\displaystyle x(x+4)=-4\)

Solve for \(\displaystyle x\).

\(\displaystyle x^2+4x=-4\)

\(\displaystyle x^2+4x+4=0\)

We can solve this either by using the quadratic formula or by factoring.  Factoring seems to be the quickest method.

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

\(\displaystyle (x+2)^2=0\)

Take the square root of both sides.

\(\displaystyle \sqrt{(x+2)^2}=\sqrt{0}\)

\(\displaystyle x+2=0\)

\(\displaystyle x=-2\)

The first integer is \(\displaystyle -2\), which we can use to find the other integer by using \(\displaystyle x+4\).

\(\displaystyle x+4=-2+4=2\)

The second integer is 2.

Therefore, the two integers are \(\displaystyle -2\) and \(\displaystyle 2\).

Since the trinomial starts with \(\displaystyle x^2\), the first term in each of the binomials will be \(\displaystyle x\):

\(\displaystyle x^2 - 3x - 10 = 0\)

\(\displaystyle (x\quad)(x\quad) = 0\)

Since you have a \(\displaystyle 10\) at the end of the trinomial, the two terms at the end of the two binomials must multiply to \(\displaystyle 10\). Since it is a negative \(\displaystyle 10\), the signs in the two binomials will be opposite.

\(\displaystyle 2\) and \(\displaystyle 5\) work:

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

Now you set each binomial equal to \(\displaystyle 0\) independently, and solve for \(\displaystyle x\).

\(\displaystyle x - 5 = 0 \qquad x + 2 = 0\)

\(\displaystyle x = 5 \qquad x= -2\)

\(\displaystyle x = 5, -2\)

\(\displaystyle \left ( x+3 \right )\left ( x+4 \right )=6\)

FOIL (first, outside, inside, last) is a device to help students remember to multiply every term by every other term exactly one time. The steps of using FOIL progress as follows:

1) Multiply the first term of each parenthetical expression together:

\(\displaystyle (x)(x)=x^2+...=6\)

2) Next, multiply the "outside" terms together:

\(\displaystyle (x)(3)=3x\)

Add that to the first product, since the terms are all being added together in the parentheses, and the distributive property requires that we maintain the sign:

\(\displaystyle x^{2}+3x...=6\)

3) Multiply the "inside" terms together:

\(\displaystyle (4)(x)=4x\)

Add that to the first two products:

\(\displaystyle x^{2}+3x+4x...=6\)

4) Finally, multiply the last terms from each parenthetical expression together:

\(\displaystyle (4)(3)=12\)

Add all of the terms together and combine like terms where possible:

\(\displaystyle x^{2}+3x+4x+12=x^{2}+7x+12=6\)

5) We now have a quadratic equation. In order to solve a quadratic, we must first set our expression equal to zero by subtracting 6 from both sides.

\(\displaystyle x^{2}+7x+6=0\)

6) Next, we factor by using backwards distribution. The idea is to split up the middle term, \(\displaystyle 7x\), in such a way that one part has a common factor with \(\displaystyle x\) and the other part has a common factor with 6. We can then factor by grouping. Obviously, no matter how we split up \(\displaystyle 7x\), one part will always contain an \(\displaystyle x\). So we must focus our attention on the 6. What are the factors of 6 that add up to 7?

\(\displaystyle 1\times 6=6\) Yes!

So,

\(\displaystyle x^{2}+x+6x+6=0\)

7) Now we pull out the common factor from each pair of terms, \(\displaystyle x\) from the first and 6 from the second:

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

8) Now we factor again, taking out the common factor \(\displaystyle \left ( x+1 \right )\) from each term:

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

9) We know that at least one of the terms must be equal to zero in order for the product to equal zero, so we set each part equal to zero and solve:

\(\displaystyle x+6=0\)

\(\displaystyle x=-6\)

and

\(\displaystyle x+1=0\)

\(\displaystyle x=-1\)

One can find the points of intersection of these two functions by setting them equal to one another, essentially substituting \(\displaystyle y\) in one equation for the \(\displaystyle x\)-side of the other equation. This will tell us when the \(\displaystyle y\) (output) will be the same in each equation for a given \(\displaystyle x\) (input).

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

By simplifying this equation and setting it equal to zero, we can find the two \(\displaystyle x\)-values that produce the same \(\displaystyle y\) values in the system of the two equations.

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

Subtract \(\displaystyle 5x\) from both sides of the equation, and add \(\displaystyle 2\) to both sides.

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

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

\(\displaystyle 6x^2 - 5x - 1 = 0\)

Factoring this last equation makes it easier to find the \(\displaystyle x\)-values that will result in zero on the left side of the equation. Set the two parenthetical phrases equal to zero to find two separate \(\displaystyle x\)-values that satisfy the equation. These \(\displaystyle x\) values will be the \(\displaystyle x\) values of the points of intersection between the two equations.

\(\displaystyle 6x^2 - 5x - 1 = 0\)

We know our factors multiply to \(\displaystyle -1\), and the six times one factor plus the other is equal to \(\displaystyle -5\)

\(\displaystyle 1*-1=-1\) and \(\displaystyle 6(-1)+1=-5\), so \(\displaystyle 1\) and \(\displaystyle -1\) are our factors.

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

\(\displaystyle 6x+1 = 0 \ \textup{and}\ x-1 = 0\)

\(\displaystyle x = -\frac{1}{6} \ \textup{and}\ x=1\)

Substituting these two values into either of the two original equations results in the \(\displaystyle y\)-values of the points of intersection.

\(\displaystyle y_1=5(1)-2=3\)

\(\displaystyle y_2=5(-\frac{1}{6})-2=-\frac{17}{6}\)

\(\displaystyle (1,3), (-\frac{1}{6}, -\frac{17}{6})\) are the points of intersection.

1) Split up the middle term so that factoring by grouping is possible.

Factors of 10 include:

1 * 10= 10    1 + 10 = 11

2 * 5 =10      2 + 5 = 7

–2 * –5 = 10    –2 + –5 = –7 Good!

\(\displaystyle x^{2}-2x-5x+10=0\)

2) Now factor by grouping, pulling "x" out of the first pair and "-5" out of the second.

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

3) Now pull out the common factor, the "(x-2)," from both terms.

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

4) Set both terms equal to zero to find the possible roots and solve using inverse operations.

x – 5 = 0,  x = 5

x – 2 = 0, x = 2