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diamong [38]
3 years ago
13

Question 10 Express 40 as a product of prime factors​

Mathematics
1 answer:
Step2247 [10]3 years ago
5 0

Answer:

2 x 2 x 2 x 5

Step-by-step explanation:

The first few prime numbers:  {2, 3, 5, 7, 11, 13, 17, 19}

One way to start is to divide by the first prime until you can't do that any more, then try the next prime and so on until all the numbers in your list are prime.

40 = 2(20)  divide 20 by 2 again

40 = 2(2)(10)  divide 10 by 2 again

40 = 2(2)(2)(5) stop--all factors are prime.

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If α and β are the zeros of the quadratic polynomial f(x) = 3x2–4x + 5, find a polynomialwhose zeros are 2α + 3β and 3α + 2β.
Lelechka [254]

Answer:

\boxed{\sf \ \ \ 3x^2-20x+37\ \ \ }

Step-by-step explanation:

Hello,

a and b are the zeros, we can say that

f(x)=3(x^2-\dfrac{4}{3}x+\dfrac{5}{3}) = 3(x-a)(x-b)=3(x-(a+b)x+ab)

So we can say that

a+b=\dfrac{4}{3}\\ab=\dfrac{5}{3}

Now, we are looking for a polynomial where zeros are 2a+3b and 3a+2b

for instance we can write

(x-2a-3b)(x-3a-2b)=x^2-(2a+3b+3a+2b)x+(2a+3b)(3a+2b)\\= x^2-5(a+b)x+6a^2+6b^2+9ab+4ab

and we can notice that

a^2+b^2=(a+b)^2-2ab so

(x-2a-3b)(x-3a-2b)=x^2-5(a+b)x+6[(a+b)2-2ab]+13ab\\= x^2-5(a+b)x+6(a+b)^2+ab

it comes

x^2-5*\dfrac{4}{3}x+6(\dfrac{4}{3})^2+\dfrac{5}{3}

multiply by 3

3x^2-20x+2*16+5=3x^2-20x+37

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