how do you factorise it?
2007-10-06 06:50:39 · 9 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Will this make it easier:
x(2x^2 - 5x - 3) ???
x(2x + 1)(x - 3)
2007-10-06 06:53:42 · answer #1 · answered by ? 7 · 4⤊ 1⤋
Well first factor out x, and get x(2x^2-5x-3). Now you have to factor 2x^2-5x+3. Because of the x^2 coefficient, the answer has to be in the form (2x+a)(x+b) for some a and b. Because of the x coefficient being negative, b is PROBABLY negative, thus a has to be positive. a and b multiply to 3. Test values for a and b: (3,-1) does not make -5 for the x-coefficient, while (1,-3) does.
You then have: x(2x+1)(x-3).
2007-10-06 06:56:27 · answer #2 · answered by vworldv 2 · 0⤊ 0⤋
(x) ( 2 x ² - 5 x - 3 )
(x) ( 2 x + 1) (x - 3 )
2007-10-06 11:05:45 · answer #3 · answered by Como 7 · 1⤊ 1⤋
2x^3 - 5x^2 - 3x
= x(2x^2 - 5x - 3)
= x(2x^2-6x + x-3)
= x[2x(x-3) + (x-3)]
= x(2x+1)(x-3)
2007-10-06 06:54:59 · answer #4 · answered by sahsjing 7 · 2⤊ 0⤋
a. The polynomial has one valuable root because of the fact there is one sign substitute. The polynomial has 2 or 0 unfavourable roots because of the fact once you negate the variables with unusual coefficients, the polynomial has 2 sign adjustments. b. the plausible rational zeros are components of persevering with/components of preferable coefficient. hence, the consistent is -4 and the preferable coefficient is two. the climate of -4, are +-a million, +- 2, and +-4. the climate of two are +-a million and +-2. the plausible rational zeros subsequently are +-a million, +-.5, +-2, and +-4.
2016-11-07 10:47:00 · answer #5 · answered by tschannen 4 · 0⤊ 0⤋
put 2x^3 - 5x^2 - 3x-o
2007-10-06 06:54:51 · answer #6 · answered by Anonymous · 0⤊ 1⤋
x(2x+1)(x-3)
2007-10-06 06:56:09 · answer #7 · answered by Laura 1 · 0⤊ 0⤋
x(2x+1)(x-3)
2007-10-06 06:56:31 · answer #8 · answered by ironduke8159 7 · 1⤊ 1⤋
Use the quadratic formula -b+/-sqrt(b^2 - 4ac)/2a, here is a website to help http://www.1728.com/quadratc.htm
2007-10-06 06:57:29 · answer #9 · answered by lance3223 1 · 0⤊ 2⤋