30b^2+21u^2-57ub stuck?

Question

factor the trinomial (hint: write the terms of the trinomial in descending powers of one variable) i dont get it..ive done plenty but i seem to be doing something wrong here..please help explain

Answer 1

When it suggests that you "write the terms of the trinomial in descending powers of one variable", it means something like this...

30b^2 - 57ub + 21u^2 (that's descending powers of b).

Straight away you can take out a factor of 3, which will make it easier...

3(10b^2 - 19ub + 7u^2)

10b^2 - 19ub + 7u^2 will factorise into something like
(?b - ?u)(?b - ?u) where you need to work out the 4 different ?

This is just like factorising a normal quadratic.

Answer 2

first write the equation like this

30b^2 - 57ub + 21u^2
next factor out the common term which in this case is 3
3[10b^2 - 19ub + 7u^]
next factor the binomial I am using trial and error.

3{(5b - 7u)(2b - 1u)]

Answer 3

30b^2 + 21u^2 - 57ub = 30b^2 - 57ub + 21u^2
powers of b go down while powers of u go up

3[10b^2 - 19ub + 7u^2] common factor

(10)(7) = 70 factors of 70 which add to -19 {-5 & -14}

3[10b^2 - 5ub - 14ub + 7u^2] re-write -19ub
3[ 5b(2b - u) - 7u(2b - u) ] common factor from each pair
3(2b - u)(5b - 7u) common factor of (2b - u)

Answer 4

it will seem complicated because the large number so try to make it smaller
divide by 3
10b^2 + 7u^2 - 19ub = 0
(2b - u)(5b - 7u) = 0
2b - u = 0 or 5b - 7u = 0
b = u/2 or b = 7u/5

Answer 5

30b^2 - b57bu + 21u^2
First factor out '3'
3[10b^2 - 19bu + 7u^2]
3[ (5b - 7u)(2b - u)] - the answer

Answer 6

30b^2+21u^2-57ub =
10b^2+7u^2-19ub = 10b^2-10ub+7u^2-7ub-2ub =
10b(b-u)+7u(u-b)-2ub = (b-u)(10b-7u) - 2ub
I think the question should be : 30b^2+21u^2-51ub

Answer 7

30b^2+21u^2-57ub=30b^2-57ub+21u^2
=3(10b^2-19ub+7u^2)
=3{10b^2-5ub-14ub+7u^2}
=3{5b(2b-u)-7u(2b-u)}
=3(2b-u)(5b-7u) ans