Question

What’s the answer?? Just a part of a homework practice

Answer 1

The functions are

`h(x)=0.42x^2+0.3x+4\text{ and }r(x)=-0.005x^2-0.2x+7`

Multiply both functions s follows.

`h(x)* r(x)=(0.42x^2+0.3x+4)*(-0.005x^2-0.2x+7)`

`=0.42x^2*(-0.005x^2-0.2x+7)+0.3x*(-0.005x^2-0.2x+7)+4*(-0.005x^2-0.2x+7)`

`=0.42x^2*(-0.005x^2)+0.42x^2*(-0.2x)+0.42x^2*7+0.3x*(-0.005x^2)+0.3x*(-0.2x)+0.3x*7+4*(-0.005x^2)+4*(-0.2x)+4*7)`

`=-0.0021x^4-0.084x^3+2.94x^2-0.0015x^3-0.06x^2+2.1x-0.02x^2-0.8x+28`

`=-0.0021x^4-0.084x^3-0.0015x^3+2.94x^2-0.06x^2-0.02x^2+2.1x-0.8x+28`

`=-0.0021x^4-0.0855x^3+2.86x^2+1.3x+28`

Hence the required product is

`q(x)=-0.0021x^4-0.0855x^3+2.86x^2+1.3x+28`

Hence the first option is correct.