How do you factor 11x^2-10x-1?
(11x + 1)(x - 1)
This one's a bit straight forward, we look at the factors of our first term's coefficient and those of our last term
so lets make two lists
11 & 1 and that's it
1 & -1 and that's it
Now we need to combine them in a way to make -10
11*1 + 1*(-1) = 10
11*(-1) + 1*1 = -10
so now we know that 11 has to multiply the -1
HI,
11x^2-10x-1=>11x^2+1-11x-1 =>x(11x+1)-(11x+1) =>(11x+1)(x-1)
11x² - 10x - 1
Polynomial like : ax² + bx + c, where :
a = 11
b = - 10
c = - 1
Δ = b² - 4ac (discriminant)
Δ = (- 10)² - 4(11 * - 1) = 100 + 44 = 144 = 12²
x1 = (- b - √Δ) / 2a = (10 - 12) / (2 * 11) = - 2/22 = - 1/11
x2 = (- b + √Δ) / 2a = (10 + 12) / (2 * 11) = 22/22 = 1
Therefore, the polynomial can be written :
= 11[x + (1/11)](x - 1)
= (11x + 1)(x - 1)
(11x+1)(x-1) would yield 11x^2 + 1x - 11x -1, which equals 11x^2 - 10x - 1
11x^2-10x-1=
=11x^2+1-11x-1=
=x(11x+1)-(11x+1)=
=(11x+1)(x-1)
Comments
(11x + 1)(x - 1)
This one's a bit straight forward, we look at the factors of our first term's coefficient and those of our last term
so lets make two lists
11 & 1 and that's it
1 & -1 and that's it
Now we need to combine them in a way to make -10
11*1 + 1*(-1) = 10
11*(-1) + 1*1 = -10
so now we know that 11 has to multiply the -1
(11x + 1)(x - 1)
HI,
11x^2-10x-1=>11x^2+1-11x-1 =>x(11x+1)-(11x+1) =>(11x+1)(x-1)
11x² - 10x - 1
Polynomial like : ax² + bx + c, where :
a = 11
b = - 10
c = - 1
Δ = b² - 4ac (discriminant)
Δ = (- 10)² - 4(11 * - 1) = 100 + 44 = 144 = 12²
x1 = (- b - √Δ) / 2a = (10 - 12) / (2 * 11) = - 2/22 = - 1/11
x2 = (- b + √Δ) / 2a = (10 + 12) / (2 * 11) = 22/22 = 1
Therefore, the polynomial can be written :
= 11[x + (1/11)](x - 1)
= (11x + 1)(x - 1)
(11x+1)(x-1) would yield 11x^2 + 1x - 11x -1, which equals 11x^2 - 10x - 1
11x^2-10x-1=
=11x^2+1-11x-1=
=x(11x+1)-(11x+1)=
=(11x+1)(x-1)