find the roots of x^3 + x^2 - 5x + 3?

Question

thanks

am trying to learn the steps to solving it. it would help me solve other questions of the same caliber.

thanks to all of you for your rapid response.. most especially puggy, you r explanation coupled with that of the book made it less complicated.

Answer 1

f (x) = x³ + x² - 5x + 3
f (1) = 1 + 1 - 5 + 3
f (1) = 0
Thus x - 1 is a root of x ³ + x ² - 5 x + 3
Find remaining roots by synthetic division:-
1 |1___1___- 5____3
_ |____1____2___- 3
_|1___2___- 3__ _0

( x - 1 ) (x ² + 2 x - 3) = 0
( x - 1 ) ( x + 3 ) ( x - 1 ) = 0
( x - 1 ) ² (x + 3) = 0
x = 1 (repeated root) , x = - 3

Answer 2

Let p(x) = x^3 + x^2 - 5x + 3

Possible rational roots: {all factors of 3} divided by {all factors of 1}. As an FYI, 3 is the number without the x, and 1 is the coefficient of the highest power of x (x^3).

Factors of 3: +1, -1, +3, -3
Factors of 1: +1, -1

Generate all factors:

{1, -1, 3, -3} / {1, -1}

Which generates

1, -1, 3, -3

Test these factors into p(x). You want a solution of 0.

p(1) = 1^3 + 1^2 - 5(1) + 3
p(1) = 1 + 1 - 5 + 3
p(1) = 2 - 2
p(1) = 0

This tells us as 1 is a root; this immediately tells us (x - 1) is a factor, so we *know* (x - 1) divides (x^3 + x^2 - 5x + 3) evenly.

From here, we use long division to solve for the remaining quadratic. But since long division is difficult to show on here, I'll show you an alternate method. I'm going to "force" (x - 1) into the equation as per the following steps.

p(x) = x^3 + x^2 - 5x + 3

Subtract and add -2x^2. Note that doing this is like adding zero, which does not change an expression.

p(x) = x^3 + x^2 - 2x^2 + 2x^2 - 5x + 3

Evaluate x^2 - 2x^2.

p(x) = x^3 - x^2 + 2x^2 - 5x + 3

Factor the first two terms.

p(x) = x^2 (x - 1) + 2x^2 - 5x + 3

Factor the last 3 terms.

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

And now, we can group (x - 1) from the entire expression.

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

And we can further factor the final quadratic.

p(x) = (x - 1)(x + 3)(x - 1)

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

We find the roots by making p(x) = 0.

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

Which means the roots are

x = { 1, -3 }

You'll notice that 1 is a repeated root.

Answer 3

You need the remainder theorem.You have a root if you can find one value for x where the value of the function(often called f(x) or y) = 0.
So I was lucky when I looked at this equation and tried x=1 which led to 1^3 + 1^2 - 5(1) +3 = 1 + 1 -5 +3 =0
So by the remainder theorem (x-1) is a factor of your equation.You now divide your equation by (x-1) to find a quadratic which I'm sure you can factorise to give the other two roots.I hope you know how to do this algebraic division because I cant figure out a way to show the method in this format.I can tell you it's like the ordinary long division method used in arithmetic.
On doing this simple division I find the quadratic to be
x^2 +2x -3 = 0 which factorises as (x +3)(x - 1) = 0
so the three roots are 1 , 1, -3

Answer 4

x³+x²-5x+3=0

The technique useful for solving other questions of this type, assuming that the factors are integers (whole numbers) is to use the REMAINDER THEOREM:

A third degree polynmial will have three integer factors whose product (all of them multiplied together ) is the constant factor 3 (the term with no x's):

Since three is a small prime, the factors are 1 * 1 * 3. with larger constant factors the combination of factors will not be so straightforward, so some experimentation will be necessary:

Substitute 1 for x: and the polynomial becomes:

1³ + 1² - 5*1 + 3 = 1 + 1 - 5 + 3 = 0 on evaluation

When the sum of all the terms is 0, then the remainder is said to be 0 and the substitued values is a factor of the polynomial.

If the remainder is not 0, then the substitued valoues is NOT a factor of the ploynomial.

factorisation continues:

x³+x²-5x+3 = (x-1)(x² + ax - 3) 'a' is the only unkown

the coefficients of x² must be equal on both sides:

x² = -x² + ax², so a = 2

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

hence x=1 (double root) and -3

Answer 5

if one of the roots is 1, the coefficients of this equation must add up to equal 0
1+1-5+3=0
that means that one of the roots is 1
plug in 1 to this equation using synthetic division
[1 1 -5 3
1[ 1 2 -3
1 2 -3 0
the equation is now:
(x-1)(x^2+2x-3)=0
factor (x^2+2x-3)=0 ----------> (x-3)(x+1)
the roots of x^3 + x^2 - 5x + 3=0 are 3,1, and -1

Answer 6

If you're ever asked this question in a HW sheet or an exam, its almost certain that the roots will be between -3 and +3 so always try those numbers to calculate the factors first, its quicker than anything else.

Answer 7

first use the factor thorem to see if you can find some factors
e.g. start with (x-1)
p(x) = x^3 + x^2 - 5x + 3
p(1) = 1 + 1 -5 + 3 = 0 so becuase there is no remainder (x-1) is a factor
now we know that there are only three factors because the highest power in this polynomial is 3
then to fin a quadratic equation we would have to divide the polynomial
x^3 + x^2 - 5x + 3 / x-1 = x^2 + 2x - 3 through long division
then factorise the quadratic
x^2 + 2x - 3 = (x + 3) (x-1) so... x=-3,1
and also there is the earlier one which is (x-1) so... x=1
so the overall answer is -3,1,1

Answer 8

1st root (of three)
x = -b/(3*a) - (2^(1/3)*(-b^2 + 3*a*c))/(3*a*(-2*b^3 + 9*a*b*c - 27*a^2*d + Sqrt[4*(-b^2 + 3*a*c)^3 + (-2*b^3 + 9*a*b*c - 27*a^2*d)^2])^(1/3)) + (-2*b^3 + 9*a*b*c - 27*a^2*d + Sqrt[4*(-b^2 + 3*a*c)^3 + (-2*b^3 + 9*a*b*c - 27*a^2*d)^2])^(1/3)/(3*2^(1/3)*a)

2nd root (of three)
x = -b/(3*a) + ((1 + i*Sqrt[3])*(-b^2 + 3*a*c))/(3*2^(2/3)*a*(-2*b^3 + 9*a*b*c - 27*a^2*d + Sqrt[4*(-b^2 + 3*a*c)^3 + (-2*b^3 + 9*a*b*c - 27*a^2*d)^2])^(1/3)) - (1 - i*Sqrt[3])*(-2*b^3 + 9*a*b*c - 27*a^2*d + Sqrt[4*(-b^2 + 3*a*c)^3 + (-2*b^3 + 9*a*b*c - 27*a^2*d)^2])^(1/3)/(6*2^(1/3)*a)

3rd root (of three)
x = -b/(3*a) + ((1 - i*Sqrt[3])*(-b^2 + 3*a*c))/(3*2^(2/3)*a*(-2*b^3 + 9*a*b*c - 27*a^2*d + Sqrt[4*(-b^2 + 3*a*c)^3 + (-2*b^3 + 9*a*b*c - 27*a^2*d)^2])^(1/3)) - (1 + i*Sqrt[3])*(-2*b^3 + 9*a*b*c - 27*a^2*d + Sqrt[4*(-b^2 + 3*a*c)^3 + (-2*b^3 + 9*a*b*c - 27*a^2*d)^2])^(1/3)/(6*2^(1/3)*a)

here are the formulas , use them , and you will get the solutions . :) hope it will help.

Answer 9

x³+x²-5x+3=0

(x+3)(x-1)(x-1)
x=-3 , 1 , 1

Answer 10

(x+3)(x+1)(x-1)

therefore x=-3 x = -1 x = 1