0
Add your answer:
Earn +20 pts
The polynomial given roots?
Do mean find the polynomial given its roots ? If so the answer is (x -r1)(x-r2)...(x-rn) where r1,r2,.. rn is the given list roots.
What is the answer to Factor r2-12r plus 27?
r^2 - 12r + 27 what two factors of 27 add up to - 12? (r - 3)(r - 9) FOIL that and get the quadratic polynomial back
How do you factor r2 plus 5r?
r(r + 5)
What the answer to factor r2 plus 12r 27?
(r + 9)(r + 3)
How do you factor r2 plus 4r plus 4?
(r + 2)(r + 2)
R2 plus 10r plus 25?
The quadratic expression: r2+10r+25 = (r+5)(r+5) when factorised
The polynomial given roots?
Do mean find the polynomial given its roots ? If so the answer is (x -r1)(x-r2)...(x-rn) where r1,r2,.. rn is the given list roots.
What is the answer to Factor r2-12r plus 27?
r^2 - 12r + 27 what two factors of 27 add up to - 12? (r - 3)(r - 9) FOIL that and get the quadratic polynomial back
How do you solve this For a polynomial of degree n express the coefficient of x to the n-1 power and the constant coefficient in terms of the zeros of the polynomial?
If the roots are r1, r2, r3, ... rn, then coeff of x^(n-1) = -(r1+r2+r3+...+rn) and constant coeff = (-1)^n*r1*r2*r3*...*rn.
Equation x2 plus Bx plus c equals 0 has 5 as the sum of its roots and 15 as the sum of the squares of its roots The value of 'c' is equals?
x2 - (-b/a)x + (c/a) = 0 or x2 - (sum of the roots)x + (product of the roots) = 0 Let the roots be r1 and r2. So we have: r1 + r2 = 5 (r1)2 + (r2)2 = 15 r1 = 5 - r2 (express r1 in term of r2) (5 - r2)2 + (r2)2 = 15 25 - 10r2 + (r2)2 + (r2)2 = 15 2(r2)2 - 10r + 25 = 15 (subtract 15 to both sides) 2(r2)2 - 10r + 10 = 0 (divide by 2 to both sides) (r2)2 - 5r + 5 = 0 (use the quadratic formula) r2 = [-b + &- sq root of (b - 4ac)]/2a r2 = {-(-5) + &- sq root of [(-5)2 - 4(1)(5)]}/2(1) = [5 + &- sq root of (25 - 20)]/2 = (5 + &- sq root of 5)/2 r1 = 5 - r2 r1 = 5 - (5 + &- sq root of 5)/2 Thus, when r2 = (5 + sq.root of 5)/2, r1 = (5 - sq.root of 5)/2 or vice versa. Since the given equation is x2 + bx + c = 0, a = 1, then c equals to the product of roots. So that, c = (r1)(r2) = [(5 - sq.root of 5)/2][(5 + sq.root of 5)/2] = [52 - (sq.root of 5)2]/4 = 5
How do you factor r2 plus 5r?
r(r + 5)
Where can you find all 25 code geass r2 in enlish?
http://dubbed-scene.com/01/c/code_geass/index_cgr2.php R2 episode 25 isn't in there yet though.
How do you factor 20r4-39r3 18r2?
r2(4r - 3)(5r - 6)
What is the answer to factor 3r5 - 192r2?
3r2(r - 4)(r2 + 4r + 16)
What the answer to factor r2 plus 12r 27?
(r + 9)(r + 3)
LCM of r2-25 and r2 plus 14r plus 45?
The LCM is r^3 + 9r^2 - 25r - 225.
Where can you find all 25 code geass r2 in English?
animefreak.tv
