Find p(D(x)) p(x)=3x^2+5x-4 , D(x)=x+3

Problem

p(D(x)),p(x)=3*x^2+5*x−4,D(x)=x+3

Solution

1. Identify the outer function p(x)=3*x^2+5*x−4 and the inner function D(x)=x+3

2. Substitute the expression for D(x) into every instance of x in the function p(x)

p*(D(x))=3*(x+3)^2+5*(x+3)−4

3. Expand the squared binomial (x+3)^2 using the formula (a+b)^2=a^2+2*a*b+b^2

p*(D(x))=3*(x^2+6*x+9)+5*(x+3)−4

4. Distribute the constants 3 and 5 into their respective parentheses.

p*(D(x))=3*x^2+18*x+27+5*x+15−4

5. Combine like terms by adding the x terms and the constant terms together.

p*(D(x))=3*x^2+23*x+38

Final Answer

p*(D(x))=3*x^2+23*x+38

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