What is the standard form of $y = {(3 x - 4)}^{2} + {(8 x - 2)}^{2}$ $y= (3x-4)^2+(8x-2)^2$ ?

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Answer 1 · 2016-01-04T09:10:25

$73 x^{2} - 56 x + 20$ $73x^2 - 56x + 20$

standard form means expanding the brackets and collecting like terms in descending order.

${(3 x - 4)}^{2} + {(8 x - 2)}^{2}$ $(3x -4 )^2 + (8x - 2 )^2$

$= (3 x - 4) (3 x - 4) + (8 x - 2) (8 x - 2)$ $= (3x - 4 )(3x - 4 ) + (8x - 2 )(8x - 2 )$

$= 3 x (3 x - 4) - 4 (3 x - 4) + 8 x (8 x - 2) - 2 (8 x - 2)$ $= 3x(3x - 4 ) -4(3x - 4 ) + 8x(8x - 2 ) -2(8x - 2 )$

$= 9 x^{2} - 12 x - 12 x + 16 + 64 x^{2} - 16 x - 16 x + 4$ $= 9x^2 - 12x - 12x + 16 + 64x^2 - 16x -16x + 4$

$= 9 x^{2} + 64 x^{2} - 12 x - 12 x - 16 x - 16 x + 16 + 4$ $= 9x^2 + 64x^2 -12x - 12x -16x - 16x + 16 + 4$

$= 73 x^{2} - 56 x + 20$ $= 73x^2 - 56x + 20$

What is the standard form of y=(3x−4)2+(8x−2)2y= (3x-4)^2+(8x-2)^2 ?