What is the standard form of $y= (3x – 4) (2x – 1) (x – 2)$ ?

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Answer 1 · 2018-03-30T10:23:16

$y=6x^3-23x^2+26x-32/3$

$y=(3x-4)(2x-1)(x-2)$

$=3xx2xx1xx(x-4/3)(x-1/2)(x-2)$

$=6(x^3-(4/3+1/2+2)x^2+(4/3xx1/2+1/2xx2+2xx4/3)x-4/3xx1/2xx2)$

$4/3+1/2+2=(8/6+3/6+12/6)=(8+3+12)/6=23/6=69/18$

$4/3xx1/2+1/2xx2+2xx4/3=2/3+1+8/3=2/3+3/3+8/3$
$=(2+3+8)/3=13/3=26/6=78/18$

$4/3xx2/3xx2=(4xx2xx2)/(3xx3)=16/9=32/18$

$(3x-4)(2x-1)(x-2)=6(x^3-69/18x^2++78/18x-32/18)$

$y=6x^3-23x^2+26x-32/3$

What is the standard form of y= (3x – 4) (2x – 1) (x – 2)y= (3x – 4) (2x – 1) (x – 2)?