Given that y = 2x^3 + 5x^2 – 8x -15,?

Given that y = 2x^3 + 5x^2 – 8x -15,?

(a) show that when x = -3, y =0;
evaluate f(- 3)...y = 2(- 27) + 5(9) + 24 - 15 = - 54 + 45 + 24 - 15 = 69 - 69 = 0
(x + 3) is a factor

(b) hence factorise 2x^3 + 5x^2 – 8x -15; and
-3]2.....5.....-8.....-15
_____-6____3___15
..2....-1......-5......0

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

(c) find the two other values of x for which y = 0.
Use the quadratic formula...

x = [1 ± √41]/4

check...do something !

y=2(-3)^3+5(-3)^2-8(-3)-15

y=-54+45+24-15=0

So (x+3) is a factor....

(2x^3+5x^2-8x-15)/(x+3)
2x^2 rem -x^2-8x-15
-x rem -5x-15
-5 rem 0

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

The other two values for x which make y=0 are

x=[1±√41]/4

y = 2x³ + 5x² - 8x - 15

y = 2x³ + (6x² - x²) - (5x + 3x) - 15

y = 2x³ + 6x² - x² - 5x - 3x - 15

y = 2x³ - x² - 5x + 6x² - 3x - 15

y = (2x³ - x² - 5x) + (6x² - 3x - 15)

y = x.(2x² - x - 5) + 3.(2x² - x - 5)

y = (x + 3).(2x² - x - 5) ← you can see that - 3 is a root


When x = - 3 → then: y = 0

The other values of x that make y = 0 are when:

2x² - x - 5 = 0

x² - (1/2).x - (5/2) = 0

x² - (1/2).x + (1/4)² - (1/4)² - (5/2) = 0

x² - (1/2).x + (1/4)² - (1/16) - (40/16) = 0

x² - (1/2).x + (1/4)² = 41/16

[x - (1/4)]² = [± (√41)/4]²

x - (1/4) = ± (√41)/4

x = (1/4) ± [(√41)/4]

x = (1 ± √41)/4

→ x₁ = (1 + √41)/4

→ x₂ = (1 - √41)/4