... x^3 - 3x - 2 look for a promptly answer to: 0 = x^3 - 3x - 2 ? 2 works (8 - 6 - 2) as a result one ingredient could desire to be (x-2). you could now do your synthetic branch. usually, you may desire to look for values which comprise ±a million, ±2, ±3, ±a million/2, and so on. If it is not obtrusive, start up homing in on a quantity via using estimates on your calculator.
Answers & Comments
Verified answer
Note that:
(3x^3 + 22x^2 + 49x + 14)/(3x + 1)
= (3x^3 + 22x^2 + 49x + 14)/[3(x + 1/3)]
= 1/3 * (3x^3 + 22x^2 + 49x + 14)/(x + 1/3).
Computing (3x^3 + 22x^2 + 49x + 14)/(x + 1/3) with synthetic division:
-1/3. . . . .|. . . . .3. . . . .22. . . . .49. . . . .14
. . . . . . . . . . . . . . . . . . -1 . . . . .-7 . . . . .-14
. . . . . . . ._ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
. . . . . . . . . . . . .3. . .. .21 . . . . .42. . . . . .0
This gives (3x^3 + 22x^2 + 49x + 14)/(x + 1/3) = 3x^2 + 21x + 42.
Thus:
(3x^3 + 22x^2 + 49x + 14)/(3x + 1)
= 1/3 * (3x^3 + 22x^2 + 49x + 14)/(x + 1/3)
= 1/3 * (3x^2 + 21x + 42)
= x^2 + 7x + 14.
I hope this helps!
... x^3 - 3x - 2 look for a promptly answer to: 0 = x^3 - 3x - 2 ? 2 works (8 - 6 - 2) as a result one ingredient could desire to be (x-2). you could now do your synthetic branch. usually, you may desire to look for values which comprise ±a million, ±2, ±3, ±a million/2, and so on. If it is not obtrusive, start up homing in on a quantity via using estimates on your calculator.
(3x+1) means x= -1/3
all exponential values are present
3,22,49,14
bring down the
3
multiply by -1/3
add to 22
21
multiply by -1/3
add to 49
42
multiply by -1/3
add to 14
0
(3x+1) is a factor
bring down the coefficients
3, 21, 42
plug in the exponents
3x^2+21+42
3(x^2+7x+6)
3(x+6)(x+1)
all together
3(3x+1)(x+6)(x+1)
x= -1/3, -6, -1
You can also use a bit of algebra to do this:
= 3x³ + 22x² + 49x + 14
= x²(3x+1) + 21x² + 49x + 14
= x²(3x+1) + 7x(3x+1) + 42x + 14
= x²(3x+1) + 7x(3x+1) + 14(3x + 1)
Now divide through by (3x+1) to get
x² + 7x + 14
I'm pretty sure you can't and you just have to use the long division method. I'm not positive though.