Can someone me with this problem asap? ?
what is the product of -2x^3+x-5 and x^3-3x-4?
show your work.
is the product of -2x^3+x-5 and x^3-3x-4 equal to the product of x^3-3x-4 and -2x^3+x-5? explain your answer.

Step-by-step explanation:

The most straightforward way to find the product is to make use of the distributive property.

  (-2x^3+x-5) · (x^3-3x-4)

  = -2x^3 · (x^3-3x-4) + x · (x^3-3x-4) - 5 · (x^3-3x-4)

  = (-2x^6 +6x^4 +8x^3) + (x^4 -3x^2 -4x) + (-5x^3 +15x +20)

  = -2x^6 +(6+1)x^4 +(8-5)x^3 -3x^2 +(-4+15)x +20

  = -2x^6 +7x^4 +3x^3 -3x^2 +11x +20

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The commutative property of multiplication applies to multiplication of polynomials. Reversing the order of the factors in the product does not change the product. The two products are the same.

Approach not taught in most schools — may require a little study and/or insight

If you study Vedic math, you find that polynomial products can be found using a sequence of X patterns on the lists of coefficients. The patterns and math are simple enough that the result can be computed in your head and you can write it down without any further ado. The coefficients of the two cubics are (highest-degree coefficient first) ...

So, the product term coefficients will be ...

-2·1, -2·0+1·0, -2·-3+1·1+0·0, -2·-4+1·-5+0·-3+0·1, 0·-4+0·-5+1·-3, 1·-4+-3·-5, -5·-4

That is, ...

-2, 0, 7, 3, -3, 11, 20 . . . . . matches the list of coefficients above for x^6 through x^0.

The idea is that the product term x^6 will be made up only of the product of the two x^3 terms.

The product term x^5 will be made up of the product of the coefficients of x^3 in the first polynomial and x^2 in the second, summed with the product of the x^2 coefficient in the first polynomial and the x^3 coefficient in the second.

The coefficient of x^4 in the product will be the sum of products of terms x^3 and x, x and x^3, x^2 and x^2.

If you locate the coefficients of these terms in the two lists above, you can see that the coefficients that get multiplied are ones at the ends of X patterns of various widths.

his chances of getting a coin of higher value is 4/19

step-by-step explanation: