What is expand ((-12+7x)(-4+5x))/(16) ?

The solution to expand ((-12+7x)(-4+5x))/(16) is (35x^2}{16}-\frac{11x)/2+3

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expand ((-12+7x)(-4+5x))/(16)

expand

\frac{( - 12 + 7 x ) ( - 4 + 5 x )}{16}

\frac{35 x ^{2}}{16} - \frac{11 x}{2} + 3

Solution steps

\frac{( - 12 + 7 x ) ( - 4 + 5 x )}{16}

Expand

(- 12 + 7 x) (- 4 + 5 x) : 48 - 88 x + 35 x^{2}

= \frac{48 - 88 x + 35 x ^{2}}{16}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{48 - 88 x + 35 x ^{2}}{16} = \frac{48}{16} - \frac{88 x}{16} + \frac{35 x ^{2}}{16}

= \frac{48}{16} - \frac{88 x}{16} + \frac{35 x ^{2}}{16}

Divide the numbers:

\frac{48}{16} = 3

= 3 - \frac{88 x}{16} + \frac{35 x ^{2}}{16}

Cancel

\frac{88 x}{16} : \frac{11 x}{2}

= 3 - \frac{11 x}{2} + \frac{35 x ^{2}}{16}

Rewrite in standard form

= \frac{35 x ^{2}}{16} - \frac{11 x}{2} + 3

(x - e)^{2} + (y - a)^{2}

n^{2} - (\frac{n ( n + 1 )}{2}) + n

(x + x^{2}) x ln (x) - x

T (- 2 (1, 0, 0) + 3 (0, 1, 0) + 1 (0, 0, 1))

i^{3}

What is expand ((-12+7x)(-4+5x))/(16) ?
The solution to expand ((-12+7x)(-4+5x))/(16) is (35x^2}{16}-\frac{11x)/2+3