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The number of items is modeled by h(x) = 0.42x2 + 0.3x + 4, and the cost per item is modeled by r(x) = −0.005x2 − 0.2x + 7. The total cost is the product of the number of items and the cost per item. Identify a polynomial q(x) that can be used to model the total cost.

Sagot :

Hello! First, let's write some important information contained in the exercise:

number of items: h(x) = 0.42x² + 0.3x + 4

cost per item: r(x) = −0.005x² − 0.2x + 7

total cost: q(x) = h(x) * r(x)

Obs: the total cost is the product of the number of items and the cost per item, so we can write it as q(x) = h(x) * r(x), right?

So, let's calculate q(x) below:

[tex]\begin{gathered} q\mleft(x\mright)=h\mleft(x\mright)\cdot r\mleft(x\mright) \\ q(x)=(0.42x^2+0.3x+4)\cdot(-0.005x^{2}-0.2x+7) \\ q(x)=0.42x^2\cdot(-0.005x^2-0.2x+7)+0.03\cdot(-0.005x^2-0.2x+7)+4\cdot(-0.005x^2-0.2x+7) \end{gathered}[/tex][tex]\begin{gathered} q(x)=0.42x^2\cdot(-0.005x^2-0.2x+7)+0.03\cdot(-0.005x^2-0.2x+7)+4\cdot(-0.005x^2-0.2x+7) \\ q(x)=-0.0021x^4-0.084x^3+0.3x+0.03\cdot(-0.005x^2-0.2x+7)+4\cdot(-0.005x^2-0.2x+7) \\ q(x)=-0.0021x^4-0.084x^3+0.3x-0.0015x^3-0.06x^2+2.1x+4\cdot(-0.005x^2-0.2x+7) \\ q(x)=-0.0021x^4-0.084x^3+0.3x-0.0015x^3-0.06x^2+2.1x-0.02x^2-0.8x+28 \end{gathered}[/tex][tex]\begin{gathered} \text{ putting similar terms in evidence:} \\ q(x)=-0.0021x^4-0.0855x^3+2.86^2+1.3x+28 \end{gathered}[/tex]

So, multiplying h(x) by r(x) we will obtain the polynomial q(x) = -0.0021x^4-0.0855x^3+2.86^2+1.3x+28

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