Change division into multiplication by the reciprocal which gives us the following

\(\displaystyle \frac{\left ( 2x-5 \right )\left ( x-3 \right )}{\left ( x-3 \right )\left ( 5-2x \right )}\)

Now \(\displaystyle \left ( 5-2x \right )=-(2x-5)\)

this results in the following:

\(\displaystyle \frac{\left ( 2x-5 \right )\left ( x-3 \right )}{\left ( x-3 \right )\left ( -1 \right )\left ( 2x-5 \right )}\)

Simplification gives us

\(\displaystyle \frac{1}{-1}\)  which equals \(\displaystyle -1\)

Here, we simply need to identify that the numerator, \(\displaystyle 3x+4\), is a factor of the denominator. Let's start by factoring \(\displaystyle 6x^{2}+5x-4\). The reverse FOIL method shows us that \(\displaystyle (3x+4)(2x-1)\) multiplies to give us \(\displaystyle 6x^{2}+5x-4\), so we can rewrite the fraction as \(\displaystyle \frac{(3x+4)}{(3x+4)(2x-1)}\). Canceling the common term gives us our answer of \(\displaystyle \frac{1}{2x-1}\).

The coefficients 16 and 64 have greatest common factor 16; there is no variable that is shared by both terms. Therefore, we can distribute out 16:

\(\displaystyle 16x^{2} + 64 = 16 \cdot x^{2 } + 16 \cdot 4 = 16 (x^{2}} + 4)\)

\(\displaystyle x^{2} + 4\) cannot be factored further, so \(\displaystyle 16 (x^{2}+4)\) is as far as we can go.

\(\displaystyle 4x^2-28 x = 0\)\(\displaystyle 4x(x-7) = 0\)\(\displaystyle x (x-7) = 0\)\(\displaystyle \fn_cm x-7 = 0 \;or\; x = 0\)

\(\displaystyle First\;equation:x-7 = 0\)

\(\displaystyle x=7\)

\(\displaystyle x=0,7\)

\(\displaystyle Find \;two \; numbers \;that \;multiply \;to \; get \;42 \; and \;add \;to \;get \; 13.\)

\(\displaystyle 6\times7=42\)

\(\displaystyle 6+7=13\)

\(\displaystyle Find\; two\; numbers\; that \; multiply \; to \; get\; -14 \; and\; add \; to\; get \; -5.\)

\(\displaystyle -7\times2=-14\)

\(\displaystyle -7+2=-5\)

First, take out the greatest common factor of the terms. The GCF of 5 and 50 is 5 and the GCF of \(\displaystyle x^{2}\) and \(\displaystyle x\) is \(\displaystyle x\), so the GCF of the terms is \(\displaystyle 5x\).

When \(\displaystyle 5x\) is distributed out, this leaves \(\displaystyle 5x (x -10)\).

\(\displaystyle x-10\) is linear and thus prime, so no further factoring can be done.

This polynomial is a difference of two squares. The below formula can be used for factoring the difference of any two squares.

\(\displaystyle a^2-b^2=(a-b)(a+b)\)

Using our given equation as \(\displaystyle \small a^2-b^2\), we can find the values to use in our factoring.

\(\displaystyle a^2-b^2=16x^2y^4-9w^4z^6\)

\(\displaystyle (4xy^2)^2=16x^2y^4\rightarrow 4xy^2=a\)

\(\displaystyle (3w^2z^3)^2=9w^4z^6\rightarrow 3w^2z^3=b\)

\(\displaystyle (a-b)(a+b)=(4xy^2-3w^2z^3)(4xy^2+3w^2z^3)\)

When factoring a polynomial that has no coefficient in front of the \(\displaystyle x^{2}\) term, you begin by looking at the last term of the polynomial, which is \(\displaystyle -35\).  You then think of all the factors of \(\displaystyle -35\) that when added together equal \(\displaystyle 2\), the coefficient in front of the \(\displaystyle x\) term.  The only combination of factors of \(\displaystyle -35\) that can satisfy this condition is \(\displaystyle -5\) and \(\displaystyle 7\).  Thus, the factors of the polynomial are \(\displaystyle (x-5)(x+7)\).

\(\displaystyle Factor \;x^2-4 x-32\)\(\displaystyle Find \; two \; numbers \; whose \; product \; is \; -32 \; and \; sum \; is -4.\)\(\displaystyle \fn_cm Those \; numbers \; are \; 4 \; and \; -8.\)
\(\displaystyle Therefore, \; x^2-4 x-32 = (x+4)(x-8)\)
\(\displaystyle Answer: \;(x+4)(x-8)\)