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Simplify the expression $\frac{5\left(\frac{x^2+3x+5}{2x-1}\right)\left(2x^2-2x-13\right)}{4x^2-4x+1}$

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Final answer to the problem

$\frac{5\left(x^2+3x+5\right)\left(2x^2-2x-13\right)}{\left(2x-1\right)\left(4x^2-4x+1\right)}$
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Step-by-step Solution

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  • Write in simplest form
  • Solve by quadratic formula (general formula)
  • Find the derivative using the definition
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1

Multiplying the fraction by $5\left(2x^2-2x-13\right)$

$\frac{\frac{5\left(x^2+3x+5\right)\left(2x^2-2x-13\right)}{2x-1}}{4x^2-4x+1}$
2

Divide fractions $\frac{\frac{5\left(x^2+3x+5\right)\left(2x^2-2x-13\right)}{2x-1}}{4x^2-4x+1}$ with Keep, Change, Flip: $\frac{a}{b}\div c=\frac{a}{b}\div\frac{c}{1}=\frac{a}{b}\times\frac{1}{c}=\frac{a}{b\cdot c}$

$\frac{5\left(x^2+3x+5\right)\left(2x^2-2x-13\right)}{\left(2x-1\right)\left(4x^2-4x+1\right)}$

Final answer to the problem

$\frac{5\left(x^2+3x+5\right)\left(2x^2-2x-13\right)}{\left(2x-1\right)\left(4x^2-4x+1\right)}$

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Function Plot

Plotting: $\frac{5\left(x^2+3x+5\right)\left(2x^2-2x-13\right)}{\left(2x-1\right)\left(4x^2-4x+1\right)}$

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7
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9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

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