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3 years ago

Factorise 9(a+b)^2 + (a+b) -8

Mathematics

1 answer:

frutty [35]3 years ago

8 0

Answer:

$(9a+9b-8)(a+b+1)$

Step-by-step explanation:

$9(a+b)^2 + (a+b)-8$

Expand $(a+b)^2$

$9(a^2+2ab+b^2)+a+b-8$

$9a^2+18ab+9b^2+a+b-8$

Now it is time to think.

Once

$9a^2=9a \cdot a$

$9b^2=9b \cdot b$

$18ab=9ab+9ab$

$-8=-8 \cdot 1$

We can get something in the format

$(9a+9b-8)(\text{some expression here})$

From the conclusions above,

$(9a+9b-8)(a+b+1)$

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3 years ago

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Step-by-step explanation:

1. Let's review the information given to us to answer the question correctly:

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3 years ago

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3 years ago

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3 years ago

Find all the complex roots. Write the answer in the indicated form. The complex cube roots of 27(cos 234° + i sin 234°) (polar f

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$\bf \textit{ roots of complex numbers, DeMoivre's theorem}
\\\\
\sqrt[n]{z}=\sqrt[n]{r}\left[ cos\left( \cfrac{\theta+2\pi k}{n} \right) +i\ sin\left( \cfrac{\theta+2\pi k}{n} \right)\right]\quad 
\begin{array}{llll}
k\ roots\\
0,1,2,3,...
\end{array}\\\\
-------------------------------$

$\bf z=27[cos(234^o)+i~sin(234^o)]
\\\\\\
\sqrt[3]{z}\implies \sqrt[3]{27}\left[ cos\left( \cfrac{234^o+360^o(0)}{3} \right) +i\ sin\left( \cfrac{234^o+360^o(0)}{3} \right)\right]
\\\\\\
3\left[ cos\left( \cfrac{234^o}{3} \right) +i\ sin\left( \cfrac{234^o}{3} \right)\right]\implies \stackrel{k=0~~1st~root}{3[cos(78^o)+i~sin(78^o)]}\\\\
-------------------------------$

$\bf \sqrt[3]{z}\implies \sqrt[3]{27}\left[ cos\left( \cfrac{234^o+360^o(1)}{3} \right) +i\ sin\left( \cfrac{234^o+360^o(1)}{3} \right)\right]
\\\\\\
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-------------------------------$

$\bf \sqrt[3]{z}\implies \sqrt[3]{27}\left[ cos\left( \cfrac{234^o+360^o(2)}{3} \right) +i\ sin\left( \cfrac{234^o+360^o(2)}{3} \right)\right]
\\\\\\
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3 0

3 years ago

Factorise 9(a+b)^2 + (a+b) -8 ​

Factorise 9(a+b)^2 + (a+b) -8