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Where P (x) is a polynomial expression then the equation (1) is known as a polynomial equation with variable x of highest power n where n = 0, 1, 2, 3, …, n.

i) Factorize the L.H.S. polynomial expression

X^2 + 8 x + 16

= x^2 + 2 * x * 4 + (4)^2

= (x + 4) ^2

ii) Now write the L.H.S. expression equal to zero then solve and find the value of x.

(x + 4) ^2 = 0

X + 4 = 0

X = - 4

Therefore x = - 4

(x – 1) (x^3 – 6 x^2 + 9 x) = 0

(x – 1) {x (x ^2 – 6 x + 9)} = 0

X (x – 1) (x ^2 – 2 * x * 3 + 3 ^2) = 0

X (x – 1) (x ^2 – 3) ^2 = 0

Therefore x = 0

X – 1 = 0, hence x = 1

(x ^2 – 3) ^2 = 0, or, x – 3 = 0, hence x = 3