Without using a calculator, determine the number of real zeros of the function f(x) = x^3 + 4x^2 + x − 6.

Question

 i been trying to do the problem forever i cant get the correct answer someone please help me with this

Answer 1

Hi Matt,

 

Have you covered synthetic division or factoring higher order polynomials?

 

If it is an x-cubed function, there is either 1 real zero or 3, and those are the only options.

 

There is always 1 real zero because the x-cubed function starts at -infinity and goes all the way to +infinity, so it MUST PASS THROUGH 0 at least once.   

 

Sometimes it bumps up and down and crosses the x-axis 3 times, but it will only look like it crosses the x-axis twice if there is a root with multiplity of 2.  Like if the x-cubed came from x(x-1)².

 

So we need to factor your polynomial, and the only rational options are ±1,2,3,6

 

Using synthetic division: 

try -1:  get a remainder of -4 so this is NOT a root

 

try -2:  get a remainder of 0, so one of your roots is -2 and it is real. 

So we have found 1 solution.  Now we need to look for others.

 

try -3:  get a remainder of 0, so one of your roots is -3 and it is real.

 

try -6:  get a remainder that is not 0, so -6 is NOT a root.

 

try 1:  get a remainder of 0, so one of your roots is 1 and it is real.

 

Since we have found 3 real roots, we do not need to look further, and the answer is 3 real roots.

 

Answer 2

f(x) = x³ + 4x² + x − 6

0 = x³ + 4x² + x − 6

 

There are always three roots (zeros) to a cubic equation, but two of them may be complex numbers rather than real numbers (provided the coefficients are rational).  So there may be either 3 or 1 real roots.  One way to solve the problem is to try factoring the cubic.  To factor you want 3 numbers that multiply to give -6 (the constant term) and add to give 4 (coefficient of the x² term).  The numbers are 3, 2, and -1: (3)(2)(-1) = -6, 3 + 2 - 1 = 4:

 

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

 

There are 3 real zeros and they are x = -3, -2, and 1