What is the standard form of $y = (x + 3) (- x - 1) + {(3 x - 7)}^{2}$ $y= (x+3)(-x-1)+(3x-7)^2$ ?

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Answer 1 · 2016-02-19T12:41:18

$8 x^{2} - 46 x + 46$ $8x^2-46x+46$

It is apparent that the highest degree of $x$ $x$ in the function

$(x + 3) (- x - 1) + {(3 x - 7)}^{2}$ $(x+3)(−x−1)+(3x−7)^2$ is two. Expanding the function

$(x + 3) (- x - 1) + (3 x - 7) (3 x - 7)$ $(x+3)(−x−1)+(3x−7)(3x-7)$

$(x + 3) (- x) - 1 (x + 3) + 3 x (3 x - 7) - 7 (3 x - 7)$ $(x+3)(−x)−1(x+3)+3x(3x-7)-7(3x-7)$ or

$- x^{2} - 3 x - x - 3 + 9 x^{2} - 21 x - 21 x + 49$ $-x^2-3x-x-3+9x^2-21x-21x+49$ or

$8 x^{2} - 46 x + 46$ $8x^2-46x+46$

As it is an function in degree $2$ $2$ of form $a x^{2} + b x + c$ $ax^2+bx+c$

What is the standard form of y= (x+3)(-x-1)+(3x-7)^2y=(x+3)(−x−1)+(3x−7)2 y= (x+3)(-x-1)+(3x-7)^2?