the real and imaginary roots of y=3x^3-7x^2-103x+35

Question

Please help

Answer 1

There are quite a few ways to tack this type of problem.

 

The quickest is to grab a graphing tool (calculator or website):

 

http://www.mathportal.org/calculators/polynomials-solvers/polynomial-graphing-calculator.php

 

This gives us the answer instantly!

 

A few other approaches involve evaluating the polynomial P(x)=y you have above for some values of x...0 and 1 make great choices!

 

Doing so yields P(0)=35 and P(1)=3-7-103+35 <0...so P has gone from a positive number to a negative one between 0 and 1!...This means there is at least one real root between x=0 and x=1...awesome!

 

Lets so we know there exists some a, 0<a<1 s.t. P(a)=0....looking at the polynomial, we see that it is of degree 3, because the highest power of x we note is x^3...so this polynomial must have a factor (x-a) and the remainder when we divide out by that factor must be a quadratic because the degrees must add.

 

So, P(x)=(x-a)(Ax^2+Bx+C), for some 0<a<1.

 

We see immediately that A=3...but lets formally show this by multiplying out and setting the coefficients of each of the four (x^k) terms when k=0,1,2,3 of P(x)...

 

Ax^3+Bx^2-aAx^2+Cx-aBx-aC = Ax^3+(B-aA)x^2+(C-aB)x-aC.

 

So A=3...we could now choose to factor the constant A into the other term, and rewrite

P(x)=(3x-3a)(x^2+bx+c)...where b=B/3 c=C/3.

 

Since the polynomial P(x) was nice (in the sense it has integer coefficients) to begin with, lets see if x=a=1/3 actually works...that is numerically the simplest choice for an a between 0 and 1 that satisfies 3x-3a=0...

 

P(1/3)=3/3^3-7/9-103/3+35=1/9-7/9-103/3+105/3=-6/9+2/3=0...yay! x=1/3 is a root!

so (3x-1) divides P(x), and the remainder is x^2+bx+c for some b,c...can we find them and apply the quadratic formula?

 

Lets multiply out (3x-1)(x^2+bx+c) and equate the coefficients:

 

3x^3+(3b-1)x^2+(3c-b)x-c=P(x)

 

so -c=35, 3c-b=-103, and 3b-1=-7, so c=-35 and b=-2 check: 3*(-35)-(-2)=? did this work?

 

Good...so P(x)=(3x-1)(x^2-2x-35)

 

Applying the quadratic formula, or simply factoring: x^2-2x-35=(x-7)(x+5) as -7 and 5 multiply to -35 and add to -2...

 

factors P(x) completely over the field of polynomials with rational coefficients...

P(x)=(3x-1)(x-7)(x+5)

 

And we can clearly see all 3 roots are real, and have values 1/3, 7, and -5.

 

Hope that helped!

 

Answer 2

When we find the roots, we let y=0.

 

 

3x³ - 7x² - 103x + 35 = 0

 

 

We can use synthetic division to find the factor of the polynomial.  This method divides the coefficients of the polynomial by the possible roots of the polynomial.  We also need a remainder of zero.  The last digit in the quotient in this method is the remainder.  We look at the last term of the polynomial.  The factors of the last term are ±1, ±5, ±7, ±35.  Lets divide the coefficients of the polynomial by -5 as a guess.

 

 

-5 |  3    -7    -103     35

 

 

______________________

 

 

 

Bring down the first coefficient below the line.

 

 
-5 |  3    -7    -103     35
 
 
______________________

       3  

 

 

 

Multiply the first digit under the line by the root.  Place the product as the addend in the second column.

 

 

 
-5 |  3    -7    -103     35
 
            -15
______________________

       3

 

 

 

Add the numbers in the second column.

 

 

 
-5 |  3    -7    -103     35
 
            -15
______________________

       3    -22

 

 

 

Repeat the above steps for the succeeding columns.

 

 

-5 |  3    -7    -103     35
 
            -15     110    -35
______________________       

       3    -22      7        0

 

 

As you can see, the last digit in the quotient is zero.  Therefore the remainder is zero.

 

The quotient is our factor.

The other factor is (x + 5).

 

 

We can write the equation as

 

(3x² - 22x + 7)(x + 5) = 0

 

 

Set the factors equal to zero.

 

3x² - 22x + 7 = 0         and           x + 5 = 0

 

 

One root is -5, just as we used in the synthetic division.

 

 

Use the quadratic formula to find the roots of the other factor.

 

 

x = (22 ± √(484 - 4(21))) / 6

 

x = (22 ± √(20)) / 6

 

x = (22 ± 2√5) / 6

 

x = (11 / 3) + (√(5) / 3)              and         x = (11 / 3) - (√(5) / 3)  

 

 

We have three real solutions.

 

x = -5

x = (11 / 3) + (√(5) / 3)             

x = (11 / 3) - (√(5) / 3)  

 

 

In addition, you may apply synthetic division a few more times to find more roots of the equation.