(2x^2+3x+4)(x^3+4x^2+3x)(x+4)
First use the distributive property for the first two:
(2x^5+8x^4+6x^3+3x^4+12x^3+9^2+4x^3+16x^2+12x)(x+4)
And then combine like terms:
(2x^5+11x^4+22x^3+25x^2+12x)(x+4)
Continue use the distributive property:
(2x^6+11x^5+22x^4+25x^3+12x^2+8x^5+44x^4+88x^3+100x^2+48x)
Continue combine like terms:
(2x^6+19x^5+66x^4+113x^3+112x^2+48x)
And that is your final answer.
Hope that help:)
Answer: wait I was wrong I saw thinking about something else
yes that is a unit rate.
Function 1:
f(x) = -x² + 8(x-15)f(x) = -x² <span>+ 8x - 120
Function 2:
</span>f(x) = -x² + 4x+1
Taking derivative will find the highest point of the parabola, since the slope of the parabola at its maximum is 0, and the derivative will allow us to find that.
Function 1 derivative: -2x + 8 ⇒ -2x + 8 = 0 ⇒ - 2x = -8 ⇒ x = -8/-2 = 4
Function 2 derivative: -2x+4 ⇒ -2x + 4 = 0 ⇒ -2x = -4 ⇒ x = -4/-2 ⇒ x= 2
Function 1: f(x) = -x² <span>+ 8x - 120 ; x = 4
f(4) = -4</span>² + 8(4) - 120 = 16 + 32 - 120 = -72
<span>
Function 2: </span>f(x) = -x²<span> + 4x+1 ; x = 2
</span>f(2) = -2² + 4(2) + 1 = 4 + 8 + 1 = 13
Function 2 has the larger maximum.
