5x2-2x+16=4x2+6x ,or x2-8x+16=0, or x2-2(x)(4)+(4)2=0, or (x-4)2=0, or (x-4)(x-4)=0, or x=4
x3 + 4x2 + 6x + 24 = (x2 + 6)(x + 4)
x2+8x = -3x2+5 4x2+8x-5 = 0 (2x+5)(2x-1) = 0 x = -5/2 or x = 1/2
Subtract 9 from each side: 4x2 + 4x - 8 = 0Simplify: 4(x2 + x - 2) = 0Factorise: 4(x + 2)(x - 1) = 0So x = -2 or 1 both of which satisfy the original equation.
4x2 -x -1 :D
x3 + 4x2 + x + 4 = (x + 4)(x2 + 1)
Given 3x3 + 4x2 +x + 7 is divided by x2 + 1, find the results:
[ x3 - 4x2 + 7x ] / (x) = x2 - 4x + 7
5x2-2x+16=4x2+6x ,or x2-8x+16=0, or x2-2(x)(4)+(4)2=0, or (x-4)2=0, or (x-4)(x-4)=0, or x=4
4x2 - 30 = 34 ⇒ 4x2 = 64 ⇒ x2 = 16 ⇒ x = 4 or -4
x3 + 4x2 + 6x + 24 = (x2 + 6)(x + 4)
x2+8x = -3x2+5 4x2+8x-5 = 0 (2x+5)(2x-1) = 0 x = -5/2 or x = 1/2
Subtract 9 from each side: 4x2 + 4x - 8 = 0Simplify: 4(x2 + x - 2) = 0Factorise: 4(x + 2)(x - 1) = 0So x = -2 or 1 both of which satisfy the original equation.
(x3 + 4x2 - 3x - 12)/(x2 - 3) = x + 4(multiply x2 - 3 by x, and subtract the product from the dividend)1. x(x2 - 3) = x3 - 3x = x3 + 0x2 - 3x2. (x3 + 4x2 - 3x - 12) - (x3 + 0x2 - 3x) = x3 + 4x2 - 3x - 12 - x3 + 3x = 4x2 - 12(multiply x2 - 3 by 4, and subtract the product from 4x2 - 12)1. 4x(x - 3) = 4x2 - 12 = 4x2 - 122. (4x2 - 12) - (4x2 - 12) = 4x2 - 12 - 4x2 + 12 = 0(remainder)
4x2-30 = 34 4x2 = 34+30 4x2 = 64 x2 = 16 x = 4
Answer : x = ± 1 4x2 - 30 = 34 : 4x2 = 34 - 30 = 4 4x2 = 4 : x2 = 1 therefore x = √1 = ± 1
4x2 -x -1 :D