Question

Given:p(x)=(2x_3)²_25

a) expand and reduce
b)factorize p(x)
c)solve p(x) =0 and p(x)=_16
d) evaluate p(5) and p(2redical 3)​

Answer 1

Explanation:

p(x) = (2x - 3)^2 - 25
a) = (2x - 3)^2 - 5^2

= (2x - 3 + 5) (2x - 3 - 5)
b) = (2x + 2) (2x - 8)

c) (2x + 2) (2x - 8) = 0 | (2x + 2) (2x - 8) = -16
4x^2 - 12x - 16 = 0 | 4x^2 - 12x - 16 = -16
x^2 - 3x - 4 = 0 | x^2 - 3x = 0
x^2 + x - 4x - 4 = 0 | x(x - 3) = 0
x(x + 1) - 4(x + 1) = 0 | x = 0 or x = 3
(x - 4) (x + 1) = 0

x = 4 or x = - 1
d) p(5) = (2(5) + 2) (2(5) - 8) | p(2root3) = (2(2root3) + 2)(2(2root3) - 8)
= 12 x 2 | = (4root3 + 2)(4root3 - 8)

= 24 | = 48 - 16 - 24root3
| = 32 - 24root3