How would one factor this expression? 28x^5 - 35x^7?

Question

What steps would one take to factor this?

Answer 1

First, take out the lowest degree of x common to all terms (in this case 5)

28x^5 - 35x^7=x^5(28-35x^2)

Then, factor out all common divisoris of the coefficients (in this case, 7)

7x^5(4-5x^2)

This is as factored as you can get.

Answer 2

pull common terms to the front: 28 and 35 are both multiples of 7, and each term has at least x^5 in it, so
28x^5 - 35x^7
7x^5[(28x^5)/(7x^5) - (35x^7)/(7x^5)]
7x^5(4 - 5x^2)
and at this point we are done for this polynomial. Technically it could be factored further down to 7(x^5)(2 +x√5)(2 - √5), but we normally don't make factorizations that force us to use radicals.

Answer 3

28x^5 - 35x^7.
Find the HCF of the numbers:
28 = 7*4, and 35 = 7*5.
HCF of 28 and 35 is therefore 7.
HCF of x^5 and x^7 is the highest power of x contained in both, namely x^5.
The common factors are therefore 7 and x^5.
Hence the result:
7x^5(4 - 5x^2).
If this were an equation with the RHS equal to 0, and you wished to solve for x, you could take the factorisation a step further, by regarding 4 - 5x^2 as the difference of two squares, and factorising that into ( 2 - sqrt(5)x )( 2 + sqrt(5)x ).
That would give final factors of:
7x^5( 2 - sqrt(5)x )( 2 + sqrt(5)x ).

Answer 4

First, factor out x^5.
(x^5)(28 - 35x^2)

Factor out -7.
-7(x^5)(5x^2 - 4)

Factor the quadratic by whatever method you prefer.
-7(5)(x^5)(x^2 - 4/5)

x² - 4/5 = 0
x² = 4/5
x = ± 2/sqrt5

-35(x^5)(x + 2/sqrt5)(x - 2/sqrt5)
That is as much as it can be factored.

Answer 5

28x^5 - 35x^7=7x^5(4-5x^2).

Answer 6

28 and 35 is divisible by 7
x^5 and x^7 s divisible by X^5
7x^5(4-5x^2)

Answer 7

take out seven x to the fifth like so:
7x^5(4-5x^2)

Answer 8

= 7x^5(4 - 5x²)
= 7x^5.(2 - √5 x).(2 + √5 x)