How do you factor: y³ - 6y² + 9y

Science & Mathematics by Anonymous 2018-06-01 12:33:37

Social Science

How do you factor: y³ - 6y² + 9y – 4 = 0?

3 answers

Wolfram tells me it's (y-4)(y-1)². How do I get there?

Anonymous

y^3-6y^2+9y-4=0 => y^3-6y^2+8y+y-4=0 => y(y^2-6y+8)+(y-4)=0 => y(y-4)(y-2)+(y-4)=0 => (y-4)[y(y-2)+1]=0 => (y-4)[y^2-2y+1]=0 => (y-4)(y-1)^2=0
Anonymous

y³ – 6y² + 9y – 4 = 0 y³ – y² – 5y² + 5y + 4y – 4 = 0 y²(y – 1) – 5y(y – 1) + 4(y – 1) = 0 (y – 1)(y² – 5y + 4) = 0 (y – 1)y – 1)(y – 4) = 0
Anonymous

y^3-6y^2+9y-4=0 Write it as y^3 -y^2 -5y^2 +5y+4y-4=0 y^2(y-1) -5y(y-1)+4(y-1)=0 (y-1)(y^2-5y+4)=0 This you can write as (y-1)(y-4)(y-1)=0 You got the solution.