Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

# Find a real root of x^3-5x^2+12x-18=0. Factorise the polynomial to show it has no other real roots.?

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• x³ - 5x² + 12x - 18 = 0

x³ - (2x² + 3x²) + (6x + 6x) - 18 = 0

x³ - 2x² - 3x² + 6x + 6x - 18 = 0

x³ - 2x² + 6x - 3x² + 6x - 18 = 0

(x³ - 2x² + 6x) - (3x² - 6x + 18) = 0

x.(x² - 2x + 6) - 3.(x² - 2x + 6) = 0

(x - 3).(x² - 2x + 6) = 0

First case: (x - 3) = 0

x - 3 = 0

x = 3

Second case: (x² - 2x + 6) = 0

x² - 2x + 6 = 0

x² - 2x = - 6

x² - 2x + 1 = - 6 + 1

(x - 1)² = - 5 ← no possible because a square is always positive or null

…so no other real root excepted 3

• f(x) = x^3 - 5x² + 12x - 18

By using the rational root theorem, the possible roots are ±1, ±2, ±3, ±6, ±9, ±18. Those are the factors of -18 divided by the factors of 1.

I'll leave some of the calculation to you, but the first one that works is x = 3.

f(3) = 3^3 - 5(3²) + 12(3) - 18

f(3) = 27 - 45 + 36 - 18

f(3) = 0

That means (x - 3) is a factor.

We can use synthetic division to divide out (x - 3):

3 | 1 . -5 . 12 . -18

...| ..... 3 .. -6 .. 18

...----------------------

.... 1 . -2 .. 6 . | . 0 <- remainder should be 0

.... ^

.... \ x² - 2x + 6

f(x) = (x - 3)(x² - 2x + 6)

You can easily show that the factored quadratic (x² - 2x + 6) has no real solutions by calculating the discriminant.

d = b² - 4ac

d = (-2)² - 4(1)(6)

d = 4 - 24

d = -20

Because the discriminant is negative, that means there are no real solutions and hence no further real roots.

f(x) = (x - 3)(x² - 2x + 6)

The only real root is x = 3

• First try the integer factors of the constant term, -18.

I usually start with the smallest absolute values: 1, then -1, then 2, then -2, etc.

Fill these in as values for x. Whichever makes the equation true is the root.

Suppose the real root is x = c. Then (x-c) is a factor.

Divide both sides of the equation by (x-c) and you'll be left with a quadratic.

Finally, use the quadratic formula to show that the other roots are complex.