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Without the requirement about integral coefficients, we could simply write (x - 1/2)(x + 1)(x - 3) = 0, but because of that extra requirement, we multiply the first factor by 2:

(2x - 1)(x + 1)(x - 3) = 0

Multiplying that out:

(2x^2 + x - 1)(x - 3) = 0

2x^3 - 5x^2 - 4x + 3 = 0

If you need more explanation, let us know.

2006-06-20 04:55:03 · answer #1 · answered by Jay H 5 · 1 0

Let f(x)=aX3(aXcube)+bX2(bXsquare)+cX+d

If 1/2,-1,3 are all roots,

it means (x-1/2) , (x+1) & (x-3) is a factor of f(x)
i.e f(1/2)=0
f(-1)=0
f(3)=0
Replacing the values in f(x)

You'll get:
f(1/2):a/8+b/4+c/2+d=0
f(-1):a+b+c+d=0
f(3):27a+9b+3c+d=0


then try to solve the eqn making 2 unknown subject of formula and using simultaneous eqn by eliminating method

2006-06-20 05:08:37 · answer #2 · answered by s13_nush 1 · 0 0

roots are 1/2, -1 and 3, so

x = 1/2
x = -1
x = 3

Transpose and multiply to get integral results
2x - 1 = 0
x + 1 = 0
x - 3= 0

Multiply
(2x - 1)(x + 1)(x - 3) = 0
(2x - 1)(x² - 2x - 3) = 0
2x³ - 5x² - 4x + 3 = 0

^_^

2006-06-21 00:31:16 · answer #3 · answered by kevin! 5 · 0 0

critical coefficients potential actual coefficients, which signifies that each and each one complicated roots ought to happen as conjugate pairs. all of us comprehend one root is -a million+i, so yet another should be -a million-i. alongside with the given -2, that provides us 3 roots, it really is sufficient to construct the cubic.

2016-11-15 00:42:43 · answer #4 · answered by ? 4 · 0 0

(x - .5)(x + 1)(x - 3) = 0 (Multiply by 2 left and right)
(2x - 1)(x + 1)(x - 3) = 0

2006-06-20 05:15:30 · answer #5 · answered by Thermo 6 · 0 0

if the roots of the equation are known...
that means
(x - root1) = 0
similarly (x - root2) = 0
and (x - root3) = 0
coz its cubic (and therefore have 3 roots)..
mutliply.. therefore for you case
(x - 1/2)(x + 1)(x - 3) = 0 is the eqaution.

2006-06-20 05:01:41 · answer #6 · answered by Varun G 3 · 0 0

(2x^2 + x - 1)(x - 3) = 0
yes got it! that can be the correct equation.

2006-06-20 07:08:07 · answer #7 · answered by ♥ Riya ♥♥♥ 2 · 0 0

(2x - 1)(x + 1)(x - 3)
(2x - 1)(x^2 - 3x + x - 3)
(2x - 1)(x^2 - 2x - 3)
2x^3 - 4x^2 - 6x - x^2 + 2x + 3
2x^3 + (-4 - 1)x^2 + (-6 + 2)x + 3
2x^3 + (-4 + (-1))x^2 + (-6 - (-2))x + 3
2x^3 - 5x^2 - 4x + 3

ANS : 2x^3 - 5x^2 - 4x + 3

2006-06-20 09:53:07 · answer #8 · answered by Sherman81 6 · 0 0

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