133k views
3 votes
Can anyone help me to solve this question ??

find the zeroes for P(x)=6x³-13x²+x+2 given 0.5 is one of the zeroes

The answers :
-1/3 , 0.5 , 2


User Jeriley
by
3.5k points

1 Answer

3 votes

Explanation:

that means the expression is the result of a multiplication with (x-0.5) being one factor.

so, we divide the whole expression by this and then deal with the remaining quadratic equation.

6x³ - 13x² + x + 2 / x - 0.5 = 6x² - 10x - 4

- 6x³ - 3x²

----------------------------

- 10x² + x + 2

- - 10x² + 5x

----------------------------

- 4x + 2

- - 4x + 2

----------------------------

finished. 0

so, the whole expression is also

P(x) = (x-0.5)×(6x² - 10x - 4)

now we need to find the 2 zero points of the quadratic part.

the general solution to a quadratic equation is

x = (-b ± sqrt(b² - 4ac))/(2a)

a = 6

b = -10

c = -4

x = (10 ± sqrt(100 - 4×6×-4))/12 =

= (10 ± sqrt(100 + 96))/12 = (10 ± sqrt(196))/12 =

= (10 ± 14)/12

x1 = (10+14)/12 = 24/12 = 2

x2 = (10-14)/12 = -4/12 = -1/3

User Bruno Martinez
by
3.3k points