Question

Hello I need help please
Factor14^{4}+ 35^{3} -7^{2}

Answer 1

\[14^{4}+ 35^{3} -7^{2} \]

Answer 2

wait its \[14x^{4}+ 35x^{3} -7x^{2}\]

Answer 3

im sorry

Answer 4

can you factor the numbers into their prime factors?

also, remember x^4 is short for x*x*x*x

the idea is to factor each term, and find things (letters or numbers) that are in each term that you can "pull out"

Answer 5

so Phi I need to find factors for 14 and 35 and 7

Answer 6

that is a good step

Answer 7

14x^4+35x^3-7x^2
\[=7x ^{2}(2x ^{2}+5x-1)\]

Answer 8

so Phi for 14 its 1& 14 , 2&7
35 is 1&35 , 5&7
and 7 is 1&7

Answer 9

Surijithayer I need a detailed report that is how I understand things better :o)

Answer 10

14 is 2*7 (though 1*14 are factors, they are not prime factors)

here is a list of the first few primes
2 3 5 7 11 13 17 19 23 29

the idea (an important idea) is any number can be broken up into multiples of *only* primes.

you get
\[ 2 \cdot 7 \cdot x \cdot x \cdot x \cdot x + 5 \cdot 7 \cdot x \cdot x \cdot x - 7 \cdot x \cdot x \]

Answer 11

now look for the same letters and numbers to "pull out"

notice all terms have a 7 and 2 x's

Answer 12

what is a prime number anyway?

Answer 13

a prime number can only be divided by itself and 1. example: 3 = 1*3 (not other factors)

not prime: 6 = 6*1 and 2*3
(2*3 is the prime factors version)

Answer 14

most people would not bother expanding the x's because you can use the exponent to know how many x's are in each term.... all terms have at least 2 x's.... so pull out x^2

Answer 15

Oh okay I get it (for both) and then what comes next(the equation)?