To find the product of (x+4) and (x^2-3x+2), you need to distribute each term in the first expression to each term in the second expression. This results in x*(x^2-3x+2) + 4*(x^2-3x+2). Simplifying further, you get x^3-3x^2+2x + 4x^2-12x+8. Combining like terms gives you the final answer: x^3+x^2-10x+8.
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Oh, what a happy little question! To find the product of (x+4) and (x^2+3x+2), we need to distribute x+4 to each term in the second expression. This gives us x^3+3x^2+2x+4x^2+12x+8. Simplifying further, we get x^3+7x^2+14x+8. Just remember, there are no mistakes, just happy little accidents!
Alright, buckle up buttercup. To multiply these two binomials, you use the distributive property twice. First, multiply x by each term in the second binomial, then do the same with 4. Finally, combine like terms to simplify the expression. It's like a math dance, but with less rhythm and more numbers.
Assuming it is x to power 4 multiplied by x to power 2 then the result is multiplied by 3xto power 2; then the result is 3 x to power 4+2+2 that equal 3 multiplied by x to power 8
(x3 + 4x2 - 3x - 12)/(x2 - 3) = x + 4(multiply x2 - 3 by x, and subtract the product from the dividend)1. x(x2 - 3) = x3 - 3x = x3 + 0x2 - 3x2. (x3 + 4x2 - 3x - 12) - (x3 + 0x2 - 3x) = x3 + 4x2 - 3x - 12 - x3 + 3x = 4x2 - 12(multiply x2 - 3 by 4, and subtract the product from 4x2 - 12)1. 4x(x - 3) = 4x2 - 12 = 4x2 - 122. (4x2 - 12) - (4x2 - 12) = 4x2 - 12 - 4x2 + 12 = 0(remainder)
The operators are missing, the possible factorisations are:4x2 + 10x + 6 = 2(2x + 3)(x + 1)4x2 - 10x + 6 = 2(2x - 3)(x - 1)4x2 + 10x - 6 = 2(2x - 1)(x + 3)4x2 - 10x - 6 = 2(2x + 1)(x - 3)
3x * (x - 4) = 3x*x - 3x*4 = 3x^2 - 12x.
3x^2
3x - x = x(3 - 1) = x(2) = 2x