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Artemon [7]
3 years ago
15

How do you factor out the Greatest Common Factor of polynomials?

Mathematics
1 answer:
In-s [12.5K]3 years ago
4 0

Answer:

Step-by-step explanation:

In order to factor out the Greatest Common Factor of polynomials, we first have to find the factors of the given polynomials. There are various methods to do that. Now we will write the steps needed to factor out the Greatest Common Factor of polynomials:

a) Breaking down every term into prime factor form.

b) Then we have to look for factors that appear in every single term in order to get Greatest Common Factor of polynomials

c) Now the Greatest Common Factor is taken out or factored out from every term before parentheses and group the remaining expression inside the parentheses.

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Read 2 more answers
URGENT HELP ME PLEASE
Trava [24]

Answer:

(a)\log_3(\dfrac{81}{3})=3

(b)\log_5(\dfrac{625}{25})=2

(c)\log_2(\dfrac{64}{8})=3

(d)\log_4(\dfrac{64}{16})=1

(e)\log_6(36^4)=8

(f)\log(100^3)=6

Step-by-step explanation:

Let as consider the given equations are \log_3(\dfrac{81}{3})=?,\log_5(\dfrac{625}{25})=?,\log_2(\dfrac{64}{8})=?,\log_4(\dfrac{64}{16})=?,\log_6(36^4)=?,\log(100^3)=?.

(a)

\log_3(\dfrac{81}{3})=\log_3(27)

\log_3(\dfrac{81}{3})=\log_3(3^3)

\log_3(\dfrac{81}{3})=3        [\because \log_aa^x=x]

(b)

\log_5(\dfrac{625}{25})=\log_5(25)

\log_5(\dfrac{625}{25})=\log_5(5^2)

\log_5(\dfrac{625}{25})=2        [\because \log_aa^x=x]

(c)

\log_2(\dfrac{64}{8})=\log_2(8)

\log_2(\dfrac{64}{8})=\log_2(2^3)

\log_2(\dfrac{64}{8})=3        [\because \log_aa^x=x]

(d)

\log_4(\dfrac{64}{16})=\log_4(4)

\log_4(\dfrac{64}{16})=1        [\because \log_aa^x=x]

(e)

\log_6(36^4)=\log_6((6^2)^4)

\log_6(36^4)=\log_6(6^8)

\log_6(36^4)=8            [\because \log_aa^x=x]

(f)

\log(100^3)=\log((10^2)^3)

\log(100^3)=\log(10^6)

\log(100^3)=6            [\because \log10^x=x]

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Answer:

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\green{ \boxed{\boxed{\begin{array}{cc}  {x}^{2}  +  {y}^{2}  = 6x - 8y \\  =  >  {x}^{2}  +  {y}^{2}   - 6x + 8y = 0 \\  =  >  {x}^{2}  +  {y}^{2}  + 2 \times ( - 3) \times x + 2 \times 4 \times y = 0 \\  \\  \sf \: standard \: equation \: o f \: circle \: is :  \\   {x}^{2}  +  {x}^{2}  + 2gx + 2fy + c = 0 \\  \\  \sf \: by \: comparing \\  \\ g =  - 3 \\ f = 4 \\ c = 0 \\  \\  \sf \: radius \:  \: r =  \sqrt{ {g}^{2}  +  {f}^{2} - c }  \\  =  \sqrt{ {( - 3)}^{2} +  {4}^{2} - 0  }  \\  =  \sqrt{9 + 16}  \\   = \sqrt{25}  \\  = 5 \: unit \\  \\  \bf \: area \:  = \pi {r}^{2}  \\  = \pi \times  {5}^{2}  \\  =\pink{ 25\pi \:  { unit }^{2} }\end{array}}}}

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