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Factorise $2x^2 + 8x + 6$ completely . Hence express 286 as the product of three prime factors .

Workings - $2x^2 + 8x + 6 = 2(x^2 + 4x + 3) =2(x+3)(x+1)$

How do I use my answer above to express 286 as the product of three prime factors ?

Thanks in advance ...

Saandeep
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user307640
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    How do you normally write $286$? What is the meaning of the location of each number? $286 = 2\cdot 10^2 + 8\cdot 10^1+6\cdot 10^0=$ two hundreds plus eight tens plus six ones. – JMoravitz Mar 19 '16 at 16:22
  • Hint; the question wants you to have $286$ equal to the expression. – lEm Mar 19 '16 at 16:23
  • $2(x+1)(x+3)=286\iff(x+1)(x+3)=143\iff(x+1)(x+3)=11\cdot13\iff(x+1)(x+3)=(10+1)(10+3)\iff{x}=10$, hence $286=2(10+1)(10+3)=2\cdot11\cdot13$. – barak manos Mar 19 '16 at 16:32

1 Answers1

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You are right- $2x^2+8x+6=2(x+3)(x+1)$

Now, $2x^2+8x+6=286$ will give you $x=10$

So, $286=2(x+3)(x+1)=2\times 13 \times 11$

The prime factors are $2,13,11$

Win Vineeth
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