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

What’s the answer?? Just a part of a homework practice-example-1
User Adriel
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1 Answer

6 votes

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.

User Ankit Bhatia
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