Chapter 6, Problem 1TP

Solution

To find which binomial expansion includes the term $40x^2$.

C. ${\left(2x+1\right)}^5$

Given information:

The term, $40x^2$.

Formula used:

The general formula of binomial expansion for $n=5$ is given as: ${\left(ax+b\right)}^2=a^5{x}^5+5\cdot a^4{x}^4\cdot b+10\cdot a^3{x}^3\cdot b^2+10\cdot a^2{x}^2\cdot b^3+ax\cdot b^4+b^5$

Calculation:

The term $40x^2$ has the exponent 2.

Observe that ${\left(ax+b\right)}^2=a^5{x}^5+5\cdot a^4{x}^4\cdot b+10\cdot a^3{x}^3\cdot b^2+10\cdot a^2{x}^2\cdot b^3+ax\cdot b^4+b^5$.

Therefore, the term containing $x^2$ is $10\cdot a^2{x}^2\cdot b^3$ or $10a^2{b}^3{x}^2$.

Now, compare each of the given binomials with ${\left(ax+b\right)}^2$ and evaluate the expression $10a^2{b}^3{x}^2$ for the values of $a$ and $b$.

For ${\left(x+1\right)}^5$ , the values are $a=1$ and $b=1$.

Then, $10a^2{b}^3{x}^2=10\cdot {\left(1\right)}^2\cdot {\left(1\right)}^3\cdot x^2=10x^2$ . That means, ${\left(x+1\right)}^5$ does not contain the term $40x^2$.

For ${\left(x+5\right)}^5$ , the values are $a=1$ and $b=5$.

Then, $10a^2{b}^3{x}^2=10\cdot {\left(1\right)}^2\cdot {\left(5\right)}^3\cdot x^2=10\cdot 125\cdot x^2=1250x^2$ . That means, ${\left(x+5\right)}^5$ does not contain the term $40x^2$.

For ${\left(2x+1\right)}^5$ , the values are $a=2$ and $b=1$.

Then, $10a^2{b}^3{x}^2=10\cdot {\left(2\right)}^2\cdot {\left(1\right)}^3\cdot x^2=10\cdot 4\cdot x^2=40x^2$ . That means, ${\left(2x+1\right)}^5$ contains the term $40x^2$.

For ${\left(2x+1\right)}^5$ , the values are $a=2$ and $b=5$.

Then, $10a^2{b}^3{x}^2=10\cdot {\left(2\right)}^2\cdot {\left(5\right)}^3\cdot x^2=10\cdot 4\cdot 125x^2=5000x^2$ . That means, ${\left(2x+5\right)}^5$ does not contain the term $40x^2$.

Conclusion:

The correct option is C. ${\left(2x+1\right)}^5$.