To divide fractions, you need to multiply the first fraction by the reciprocal of the second.

\(\displaystyle (\frac{12m^2n}{16m^3n^4})\div(\frac{8m^{-4}n}{18m^4n^2})=(\frac{12m^2n}{16m^3n^4})\times(\frac{18m^4n^2}{8m^{-4}n})\)

Now, multiply the numerators and denominators, then simplify.

\(\displaystyle (\frac{12m^2n}{16m^3n^4})\times(\frac{18m^4n^2}{8m^{-4}n})=\frac{216m^6n^3}{128m^{-1}n^5}=\frac{27m^7}{16n^2}\)

Cross multiply to eliminate the fractions.

\(\displaystyle (1)\sqrt{3x+1} = (1)(3)\)

\(\displaystyle \sqrt{3x+1} =3\)

Square both sides.

\(\displaystyle (\sqrt{3x+1})^2 =3^2\)

\(\displaystyle 3x+1 =9\)

Subtract 1 on both sides.

\(\displaystyle 3x+1 -1=9-1\)

\(\displaystyle 3x=8\)

Divide by 3 on both sides.

\(\displaystyle \frac{3x}{3}=\frac{8}{3}\)

The answer is:  \(\displaystyle \frac{8}{3}\)

When the baseball hits the ground, the height is 0, so we set \(\displaystyle h\left ( t\right ) = 0\). and solve for \(\displaystyle t\).

\(\displaystyle -16t^{2}+50t + 120 = 0\)

This can be done using the quadratic formula:

\(\displaystyle t = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\)

Set \(\displaystyle a = -16, b = 50, c = 120\):

\(\displaystyle t = \frac{-50 \pm \sqrt{50^{2}-4(-16)120}}{2(-16)}\)

\(\displaystyle t = \frac{-50 \pm \sqrt{2,500- (- 7,680)}}{-32}\)

\(\displaystyle t = \frac{-50 \pm \sqrt{2,500+7,680}}{-32}\)

\(\displaystyle t = \frac{-50 \pm \sqrt{10,180}}{-32}\)

\(\displaystyle t \approx \frac{-50 \pm 100.9}{-32}\)

One possible solution:

\(\displaystyle t \approx \frac{-50 +100.9}{-32} \approx -1.6\) 

We throw this out, since time must be positive.

The other:

\(\displaystyle t \approx \frac{-50 -100.9}{-32} \approx 4.7\)

This solution, we keep. The baseball hits the ground in about 4.7 seconds.

By definition, \(\displaystyle (g \circ f) (x) = g (f(x))\), so

\(\displaystyle (g \circ f) (x) = g ( x^{2} - 5 )\)

\(\displaystyle = 2 (x^{2}-5 ) + 7\)

\(\displaystyle = 2 x^{2}-10 + 7\)

\(\displaystyle = 2 x^{2}-3\)

Factoring a quadratic equation means doing FOIL backwards. Recall that when you use FOIL, you start with two binomials and end with a trinomial:

\(\displaystyle (x+a)(x+b)=x^{2}+(a+b)x+ab\)

Now, we're trying to go the other direction -- starting with a trinomial, and going back to two factors.

Here, -3 is equal to \(\displaystyle ab\), and -2 is equal to \(\displaystyle a+b\). We can use this information to find out what \(\displaystyle a\) and \(\displaystyle b\) are, separately. In other words, we have to find two factors of -3 that add up to -2.

Factors of -3:

• 3*-1 (sum = 2)
• -3*1 (sum = -2)

Thus our factored equation should look like this:

\(\displaystyle y=(x-3)(x+1)\)

The roots of the quadratic equation are the values of x for which y is 0.

\(\displaystyle 0=(x-3)(x+1)\)

We know that anything times zero is zero. So the entire expression equals zero when at least one of the factors equals zero.

\(\displaystyle x-3=0\: \: \textup{and} \: \: x+1=0\)

\(\displaystyle x=3 \: \: \textup{and}\: \: -1\)

Factor:

\(\displaystyle 0=x^2+4x-12\)

\(\displaystyle \small \small 0=(x-2)(x+6)\)

Double check by factoring:

\(\displaystyle \small F: (x)(x)\)

\(\displaystyle \small \small O:(x)(6)=6x\)

\(\displaystyle \small \small I: (-2)(x)=-2x\)

\(\displaystyle \small \small L: (-2)(6)=-12\)

Add together: \(\displaystyle \small x^2+6x-2x-12=x^2+4x-12\)

Therefore:

\(\displaystyle \small x-2=0\ or x+6=0\)

\(\displaystyle \small x=2\or-6\)

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

\(\displaystyle x^{2}-5x+9=-13x-7\)

1) First step of solving any equation: combine like terms. With quadratics, the easiest step to take is to set the expression equal to zero.

\(\displaystyle x^{2}+8x+16=0\)

2) There are two ways to do this problem. The first and most intuitive method is standard factoring.

16 + 1 = 17

8 + 2 = 10

4 + 4 = 8

\(\displaystyle x^{2}+4x+4x+16=0\)

3) Then follow the usual steps, pulling out the common factor from both pairs, "x" from the first and "4" from the second.

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

4) Pull out the "(x+4)" to wind up with:

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

5) Set each term equal to zero.

x + 4 = 0, x = –4

But there's a shortcut! Assuming the terms are arranged by descending degree (i.e., \(\displaystyle px^{2}+qx+k=0\)), and the third term is both a perfect square whose square root is equal to half of the middle term, mathematicians use a little trick. In this case, the square root of 16 is 4. 4 * 2=8, so the trick will work. Take the square root of the first and last term, then stick a plus sign in between them and square the parentheses. 

\(\displaystyle \sqrt{x^{2}}=\left | x\right |=x\)

\(\displaystyle \sqrt{16}=4\)

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

And x, once again, is equal to –4.

To solve for \(\displaystyle x\), you need to isolate it to one side of the equation. You can subtract the \(\displaystyle 6x\) from the right to the left. Then you can add the 6 from the right to the left:

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

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

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

Next, you can factor out this quadratic equation to solve for \(\displaystyle x\). You need to determine which factors of 8 add up to negative 6:

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

\(\displaystyle (x-2)(x-4)=0\)

Finally, you set each binomial equal to 0 and solve for \(\displaystyle x\):

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

Add 8 to both sides to set the equation equal to 0:

\(\displaystyle \small 3h^2+10h+8=0\)

To factor, find two integers that multiply to 24 and add to 10.  4 and 6 satisfy both conditions. Thus, we can rewrite the quadratic of three terms as a quadratic of four terms, using the the two integers we just found to split the middle coefficient:

\(\displaystyle \small 3h^2+6h+4h+8=0\)

Then factor by grouping:

\(\displaystyle \small 3h(h+2)+4(h+2)=0 \rightarrow (h+2)(3h+4)=0\)

Set each factor equal to 0 and solve:

\(\displaystyle h=-2\)

and

\(\displaystyle h=-\frac{4}{3}\)