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Question

1. One factor of x^3+x^2 - 14x - 24 is x -4 find the remaining factors

 

2.  state the possible number of positive real zeroes, negative real zeroes and imaginary real zeroes for g(x) = 3x^5 - 2x^3-4x^2+8x-1

 

3. Find all the possible zeroes of the function, then find the zeroes of f(x) = x^3-5x^2 + 8x-6

 

4. Write a polynomial function of least degree with integral coefficients whose zeroes include 4 and 1 - i

 

Thanks in advance!

Answer 1

Problem 1:

 

x³ + x² - 14x - 24

 

= x³ - 4x² + 5x² - 20x + 6x - 24

 

= x²(x - 4) + 5x(x - 4) + 6(x - 4)

 

= (x² + 5x + 6)(x - 4)

 

= (x + 2)(x + 3)(x - 4)

 

So the other factors are x+2 and x+3.

 

 

Problem 2:

 

Use Descartes' rule of signs.

 

3x⁵ - 2x³ - 4x² + 8x - 1

 

Three sign flips between consecutive terms.

Thus there are either 1 or 3 positive real zeroes .

 

 

If we multiply the odd degree coefficients by -1, we get -3,2,-4,-8,-1, with two sign flips.

Thus there are either 0 or 2 negative real zeroes.

 

Since we from the above, we may have all 5 roots be real, there can be 0, 2, or 4 imaginary roots.

 

Problem 3:

x³ - 5x² + 8x - 6

 

Possible rational roots: ±(factor of -6)/(factor of 1)

 

±1, ±2, ±3, ±6

 

Substituting shows that 3 is the only rational root.

 

Factor out x - 3.

 

x³ - 5x² + 8x - 6

 

= x³ - 3x² - 2x² + 6x + 2x - 6

 

= x²(x - 3) - 2x(x - 3) + 2(x - 3)

 

= (x² - 2x + 2)(x - 3)

 

Solve for the other two roots.

 

x² - 2x + 2 = 0

 

x² - 2x + 1 = -1

 

x - 1 = ±i

 

x = 1 ± i

 

Problem 3:

 

For any polynomial with real coefficients, complex roots come in conjugate pairs, so 1+i is also a root.

 

From the last calculations above, 1 + i and 1 - i are roots of x² - 2x + 2. The minimum polynomial with integer coefficients with 4 and 1 - i as roots is therefore

 

(x - 4)(x² - 2x + 2) = x³ - 6x² + 10x - 8