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find the zero of the following polynomial p(x) = (x+2) (2x²+3x-9)A) x= -3,-2,2/3B) x= -3,-2C) x=-3,-2,3/2D) x= 3/2,2,3

User Andrey Sboev
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1 Answer

14 votes
14 votes

x=-3,-2,3/2 (option C)

Step-by-step explanation:

p(x) = (x+2) (2x²+3x-9)

to get the zeros, we equatoe the factors with zero

x + 2 = 0

x = -2

2x²+3x-9 = 0

we factorise using factorisation method

a = 2, b= 3, c = -9

The product of a and c = -18

two factors of -18 that sum up to 3 but their product gives -18 is 6 and -3

2x²+6x-3x-9 = 0

2x(x + 3) -3(x + 3) = 0

(2x-3)(x+3) = 0

2x-3 = 0 or x+3 = 0

2x = 3 or x = -3

x = 3/2 or x = -3

Hence, the factors of p(x) = (x+2) (2x²+3x-9) are x = -2, 3/2, -3

The correct option: x=-3,-2,3/2 (option C)

User Kevin Brady
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