Question

1 Using factor theorem, factorise x3-6x2+3x+10.
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Answer 1

Step-by-step explanation:

Let f(x) = x^3 – 6x^2 + 3x + 10

Constant term = 10

Factors of 10 are ±1, ±2, ±5, ±10  

Let x + 1 = 0 or x = -1  

f(-1) = (-1)^3 – 6(-1)^2 + 3(-1) + 10  

= 10 – 10

= 0  

f(-1) = 0  

Let x + 2 = 0 or x = -2  

f(-2) = (-2)^3 – 6(-2)^2 + 3(-2) + 10  

= -8 – 24 – 6 + 10  

= -28  

f(-2) ≠ 0  

Let x – 2 = 0 or x = 2  

f(2) = (2)^3 – 6(2)^2 + 3(2) + 10  

= 8 – 24 + 6 + 10

= 0  

f(2) = 0  

Let x – 5 = 0 or x = 5  

f(5) = (5)^3 – 6(5)^2 + 3(5) + 10

= 125 – 150 + 15 + 10

= 0  

f(5) = 0

Therefore, (x + 1), (x – 2) and (x - 5) are factors of f(x)  

Hence f(x) = (x + 1) (x – 2) (x - 5)

Answer 2

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