What is the product of (6x3+3x)(x2+4)

1 answer:

3 0

If you would like to know what is the product of (6 * x^3 + 3 * x) * (x^2 + 4), you can calculate this using the following steps:

(6 * x^3 + 3 * x) * (x^2 + 4) = 6 * x^3 * x^2 + 6 * 4 * x^3 + 3 * x * x^2 + 3 * 4 * x = 6 * x^5 + 24 * x^3 + 3 * x^3 + 12 * x = 6 * x^5 + 27 * x^3 + 12 * x

The correct result would be 6 * x^5 + 27 * x^3 + 12 * x.

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Ronch [10]

Consider all options:

$\left[\begin{array}{cc}5&5\\-4&-4\end{array}\right] \cdot \left[\begin{array}{cc}5&5\\-4&-4\end{array}\right] =\left[\begin{array}{cc}5\cdot 5+5\cdot (-4)&5\cdot 5+5\cdot (-4)\\-4\cdot 5+(-4)\cdot (-4)&-4\cdot 5+(-4)\cdot (-4)\end{array}\right]=$

$=\left[\begin{array}{cc}5&5\\-4&-4\end{array}\right].$

This option is true.

$\left[\begin{array}{cc}6&5\\5&6\end{array}\right] \cdot \left[\begin{array}{cc}6&5\\5&6\end{array}\right] =\left[\begin{array}{cc}6\cdot 6+5\cdot 5&6\cdot 5+5\cdot 6\\5\cdot 6+6\cdot 5&5\cdot 5+6\cdot 6\end{array}\right]=$

$=\left[\begin{array}{cc}61&60\\60&61\end{array}\right].$

This option is false.

$\left[\begin{array}{cc}0.5&-0.5\\-0.5&0.5\end{array}\right] \cdot \left[\begin{array}{cc}0.5&-0.5\\-0.5&0.5\end{array}\right] =\left[\begin{array}{cc}0.5\cdot 0.5+(-0.5)\cdot (-0.5)&0.5\cdot (-0.5)+(-0.5)\cdot 0.5\\-0.5\cdot 0.5+0.5\cdot (-0.5)&-0.5\cdot (-0.5)+0.5\cdot 0.5\end{array}\right]=$

$=\left[\begin{array}{cc}0.5&-0.5\\-0.5&0.5\end{array}\right].$

This option is true.

$\left[\begin{array}{cc}0.5&0.5\\-0.5&0.5\end{array}\right] \cdot \left[\begin{array}{cc}0.5&0.5\\-0.5&0.5\end{array}\right] =\left[\begin{array}{cc}0.5\cdot 0.5+0.5\cdot (-0.5)&0.5\cdot 0.5+0.5\cdot 0.5\\-0.5\cdot 0.5+0.5\cdot (-0.5)&-0.5\cdot 0.5+0.5\cdot 0.5\end{array}\right]=$

$=\left[\begin{array}{cc}0&0.5\\-0.5&0\end{array}\right].$

This option is false.

$\left[\begin{array}{cc}-6&-6\\5&5\end{array}\right] \cdot \left[\begin{array}{cc}-6&-6\\5&5\end{array}\right] =\left[\begin{array}{cc}-6\cdot (-6)+(-6)\cdot 5&-6\cdot (-6)+(-6)\cdot 5\\5\cdot (-6)+5\cdot 5&5\cdot (-6)+5\cdot 5\end{array}\right]=$