divide and simplify
1) d+2 7d+14
___ / ______
d-8 d-3
2) v^2-64 v-8
_____ / ___
81v+648 72
3) 3x^2-23x+14 21x^2-20x+4
__________ / __________
3x^2+7x-6 7x^2+19x-6
2007-09-02 16:28:18 · 5 answers · asked by Lucy 1 in Science & Mathematics ➔ Mathematics
Hi,
1) Change from division to multiplying by the reciprocal of the second fraction. Then factor and cancel out common factors.
d+2.....7d+14
-----.÷.----------
d-8...... d-3
d+2.....d-3
-----.X.---------- =
d-8......7d+14
d+2.....d-3
-----.X.---------- =
d-8......7(d+2)
Cancel out the d+2 factor.
d-3
-------- <== answer
7(d-8)
2) Change from division to multiplying by the reciprocal of the second fraction. Then factor and cancel out common factors.
v²-64..... .....v-8
-------------.÷.------
81v+648......72
v²-64..... .....72
-------------.X.------
81v+648......v-8
Now factor and cancel.
(v-8)(v+8).......72
-------------.X.------
81(v+8).........v-8
Cancel out the v+8 factors and the v-8 factors. Then reduce the fraction to lowest terms.
72....8
--- = --- <== answer
81.....9
3) Change from division to multiplying by the reciprocal of the second fraction. Then factor and cancel out common factors.
3x^2-23x+14....21x^2-20x+4
------------------.÷.-------------------
3x^2+7x-6......... 7x^2+19x-6
3x²-23x+14....7x²+19x-6
----------------.X.---------------
3x²+7x-6........ 21x²-20x+4
Now factor each polynomial.
(x-7)(3x-2)......(x+3)(7x-2)
----------------.X.--------------- =
(x+3)(3x-2)......(3x-2)(7x-2)
Cancel out the 3x - 2 factors, the x+3 factors, and 7x-2 factors.
x-7
------ <== answer
3x-2
I hope these help you!! :-)
2007-09-02 16:59:10 · answer #1 · answered by Pi R Squared 7 · 0⤊ 0⤋
1)
(d + 2/d - 8)/(7d + 14)/(d - 3) =
(d + 2)((d - 3)/(d - 8)(7d + 14) =
(d + 2)(d - 3)/7(d + 2)(d - 8) =
(d - 3)/7(d - 8)
2)
[(v^2 - 64)/(81v + 648)]/(v - 8)(72) =
[(72)(v^2 - 64)]/(81v + 648)(v - 8) =
72(v + 8)(v - 8)/81(v + 8)(v - 8) =
72/81 = 8/9
3)
[(3x^2 - 23x + 14)/(3x^2 + 7x - 6)]/(21x^2 - 20x + 4)/(7x^2 + 19x - 6)
factorising each expression
3x^2 - 23x + 14 = 3x^2 - 21x - 2x + 14 =
3x(x - 7) - 2(x - 7) = (x - 7)(3x - 2)
3x^2 + 7x - 6 = 3x^2 + 9x -2x - 6 =
3x(x + 3) - 2(x + 3) =
(x+ 3)(3x - 2)
21x^2 - 20x + 4 = 21x^2 - 14x - 6x + 4 =
7x(3x - 2) - 2(3x - 2) =
(3x - 2)(7x - 2)
7x^2 + 19x - 6 = 7x^2 + 21x - 2x - 6 =
7x(x + 3) - 2(x + 3) =
(x + 3)(7x - 2)
so
[(3x^2 - 23x + 14)/(3x^2 + 7x - 6)]/(21x^2 - 20x + 4)/(7x^2 + 19x - 6) =
[(x - 7)(3x - 2)/(x + 3)(3x - 2)]/(3x - 2)(7x - 2)/(x + 3)(7x - 2) =
[(x - 7)/(x + 3)] /[(3x - 2)/( x + 3)] =
(x - 7)(x + 3)/(x + 3)(3x - 2) = (x - 7)/(3x - 2)
2007-09-03 00:07:27 · answer #2 · answered by mohanrao d 7 · 0⤊ 0⤋
Well, we none of us are going to do all of your problems.... What you need to do with this set of stuff is get it factored so it looks like ( ) ( ) all over ( ) ( ).
Then you can cancel.
On some of these, you have to collect like terms before you can factor them.... as in the numerator of number 3
collect 3x^2 and 21x^2 etc before you try to factor. And they will factor into ( )(). That's why you were given these exercises.... to practice factoring, and canceling
2007-09-02 23:37:10 · answer #3 · answered by April 6 · 0⤊ 0⤋
Write your problems so they are easily understood.
2007-09-02 23:34:40 · answer #4 · answered by ironduke8159 7 · 0⤊ 0⤋
Please write your question clearer. It cannot be understood.
2007-09-02 23:47:16 · answer #5 · answered by Anonymous · 0⤊ 0⤋