Maths .......HELP?

Question

1. What is the quotient of y-5/ 2y^2-7y-15?

2. What is the quotient of b^3+4b^2-3b+126/ b+7?

3. What is the quotient of d-2/ d^4-6d^3+d+17?

THANKS

Answer 1

1. You need to factor the larger expression. When you do, you will see that one of the factors happens to be (y-5), so they have made this one pretty easy for you.

Do you know how to factor? you will get (y-5) * (2y + 3) for the 2 factors of 2y^2 -7y -15. From there the division is easy, as one term cancels out, the (y-5).

for 2 and 3, you must have learned in class EITHER how to factor a cubic like number 2 or a quartic like number 3, in which case it would be the very same approach. (guessing now, but I anticipate that (b+7) and (d-2) respectively will be one of the factors in 2 and in 3, but maybe not in 3.)

Another way to do it if that is the case is to ask what times b+7 would = b^3 +4b^2 -3b +126? (hint- it HAS to start with b^2 so you get the b^3 term, and END with +18 so the 7x18=126)

It's POSSIBLE you have learned how to actually DIVIDE polynomials using long division. In that case you set it up just like any long division problem, asking first how many times does b go into b^3? answer.... b^2.... then write b^2 above the long division mark and multiply b^2 times b and times 7... getting b^3 + 7b^2, which you write underneath as in any long division problem and subtract.


it will all look something like this:


.........b^2
...... ____________________
b+7| b^3 + 4b^2 - 3b + 126
........b^3 + 7b^2
.......-----------------
subtract......-3b^2 - 3b (bringing down the -3b) continue,

asking how many times b+7 goes into -3b^2 -3b.... etc.

hope that helps either way. :-)

Answer 2

Too many questions.
I assume you left out some parentheses.

The principle is: Factorise both top and bottom, and cancel out common factors. I'll so question 2 and start you off on 3.

Since b+7 is the denominator, I'm hoping it's a factor of the top. And it is, because when you sub -7 for b in the top you get 0. Can you do synthetic division?
It gives (b^2-3b+18), which is therefore the required quotient.

In question 3, obviously d-2 is not a factor of the bottom because 2 is not a factor of 17. So is this the wrong way round? Are you supposed to divide d^4-6d^3+d+17 by d-2?

If so, you should get remainder -13, I think.

1.......-6.......0.......1.....17
....... .2.......-8.....-16...-30
1......-4.......-8.....-15...-13

That makes the quotient d^3-4d^2-8d-15 with remainder -13

Answer 3

1.(y-5)/(2y^2-7y-15)
=(y-5)/2y^2-10y+3y-15)
=(y-5)/{2y(y-5)+3(y-5)
=(y-5)/(y-5)(2y+3)
=1/(2y+3) [after cancelling y-5 of numerator and denomior]
2.b^3+4b^2-3b+126/(b+7)
={b^2(b+7)-3b(b+7)+18(b+7)}/(b+7)
=(b+7)(b^2-3b+18)/(b+7)
=b^2-3b+18 [after cancelling b+7 of numerator and denomiator]
3.(d-2)/(d^4-6d^3+d+17)
=(d-2)/{d^3(d-2)-4d^2(d-2)-8d(d-2)-15(d-2)-13
Sorry student,,there is some mistake in the problem as some remainder is coming.However,if we donot take the remainder in consideration,then the quotient will be1/(d^3-4d^2-8d-15)
Will you please check up the problem once more and if there is any mistake found by you,please send it by a mail to me through Yahoo Answer and I promise to send it back properly solved.

Answer 4

1/y+3

y-5/ (y+3)(y-5)

(cancel)

1/y+3!!!!!!!

you can apply this on the others.....

Answer 5

1 + 1 = WINDOW!!!

MY MUMMY SAYS IM SMART!!!

Answer 6

work it out yourself!(or ask jeeves!)