this question makes no sense. do you mean what are the zeros/roots of this polynomial?
Factoring trinomials are fun!
5x2 - 8x - 4
(5x + 2)(x - 2)
Never forget to check:
5x2 - 10x + 2x - 4
5x2 - 8x - 4
Good everything is in order, now set each parentheses equal to zero to solve for x
5x + 2 = 0
5x = -2
x = - 2/5
x - 2 = 0
x = 2
All done! Now we know the solution set is {-2/5, 2}. Remember smallest number always goes first.
5x2-8x-4
If you notice, 8x + 4 = 4(2x + 1), but there is an odd coefficient for x2, so there is guaranteed to be some remainder (that is 8x + 4 is not a factor of the polynomial): (24x3 - 5x2 - 48x - 8) ÷ (8x + 4) = 3x2 - 2x - 5 - (x2 - 12)/(8x + 4)
In algebra and mathematics the simple above equation can be written and simplified as follows . (5x2 8x)-(3x2-x) = 10 + 8x -6 +x = 4 +7x.
(x - 2)(x - 2)(x - 1)
5x2 + 8x = 7 5x2 + 8x - 7 = 0 This cannot be factorised so the solutions need to be determined using the quadratic formula The solutions are {-8 ± sqrt[82 - 4*5*(-7)]}/(2*5) = {-8 ± sqrt[64 + 140]}/10 = {-8 ± sqrt[204]}/10 = -2.22829 and 0.62829 (to 5 dp)
5x2-8x-4
If you notice, 8x + 4 = 4(2x + 1), but there is an odd coefficient for x2, so there is guaranteed to be some remainder (that is 8x + 4 is not a factor of the polynomial): (24x3 - 5x2 - 48x - 8) ÷ (8x + 4) = 3x2 - 2x - 5 - (x2 - 12)/(8x + 4)
In algebra and mathematics the simple above equation can be written and simplified as follows . (5x2 8x)-(3x2-x) = 10 + 8x -6 +x = 4 +7x.
(5x+2) and (x-2)
(x - 2)(x - 2)(x - 1)
5x2 + 8x = 7 5x2 + 8x - 7 = 0 This cannot be factorised so the solutions need to be determined using the quadratic formula The solutions are {-8 ± sqrt[82 - 4*5*(-7)]}/(2*5) = {-8 ± sqrt[64 + 140]}/10 = {-8 ± sqrt[204]}/10 = -2.22829 and 0.62829 (to 5 dp)
(4x2) (x/4)(8x - 32) =x3(8x - 32) = 8x3(x - 4)
(5x - 7)(x + 3)
5x2=8x Divide with x on both sides5x= 8 Solve xx=8/5
To factorise x3 + 5x2 - 16x - 80 I note that -80 = 5 x -16 and 5 & -16 are the coefficients of x2 and x. Thus I have: x3 + 5x2 - 16x - 80 = (x + 5)(x2 - 16) and the second term is a difference of 2 squares, meaning I have: x3 + 5x2 - 16x - 80 = (x + 5)(x + 4)(x - 4)
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
(4x-8)(2x+2)