Question

show that X= 2 is a root of 2x^3+x^2-7x-6

Answer 1

Step-by-step explanation:

2x³ + x² - 7x -6 = 0

2x³ + 2x² - x² - x - 6x - 6 = 0

2x²(x + 1) -x(x + 1) -6(x + 1) = 0

(x + 1)(2x² - x - 6) = 0

(x + 1) ( 2x² - 4x + 3x - 6) = 0

(x + 1) (2x(x -2) + 3(x - 2)) = 0

(x + 1)(2x + 3)(x - 2) = 0

x = -1, x = -3/2, x = 2

Answer 2

⏺️p(x) = 2x³ + x² -7x +6

We know ,

By remainder theorem,

If p(2) is 0 , then

X = 2 is a root of p(x)

⏺️p(2) = 2(2)³ + (2)² -7(2) +6 = 16+4-14-6 = 0

Hence , X= 2 is a root of the Polynomial.