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

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

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Answer 1 · 2016-02-17T15:18:35

Multiply long hand simplify to get:
$3 x^{2} - 2 x - 16$ $3x^2-2x-16$

Answer 2 · 2016-02-17T15:41:56

$3 x^{2} - 2 x - 16$ $3x^2 - 2x - 16$

Explanation:

distribute the 1st 'pair' of brackets using FOIL (or any method you use).

(x-3)(4x+8) = $4 x^{2} + 8 x - 12 x - 24 = 4 x^{2} - 4 x - 24$ $4x^2 + 8x -12x -24 = 4x^2 -4x - 24$

Similarly for the 2nd 'pair' of brackets.

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

now have : $4 x^{2} - 4 x - 24 - (x^{2} - 2 x - 8)$ $4x^2 - 4x - 24 - ( x^2 - 2x - 8 )$

$= 4 x^{2} - 4 x - 24 - x^{2} + 2 x + 8 = 3 x^{2} - 2 x - 16$ $= 4x^2 - 4x - 24 - x^2 + 2x + 8 = 3x^2 - 2x - 16$

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

2 Answers

Explanation:

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

2 Answers

Explanation:

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