SATMathematics Practice Test 1

One can find the possible values of a and b in $\left(\mathit{a}\mathit{x}+2\right)\left(\mathit{b}\mathit{x}+7\right)$ by using the given equation a + b = 8 and finding another equation that relates the variables a and b.
Since $\left(\mathit{a}\mathit{x}+2\right)\left(\mathit{b}\mathit{x}+7\right)=15{\mathit{x}}^2+\mathit{c}\mathit{x}+14$, one can expand the left side of the equation to obtain $\mathit{a}\mathit{b}{\mathit{x}}^2+7\mathit{a}\mathit{x}+2\mathit{b}\mathit{x}+14=15{\mathit{x}}^2+\mathit{c}\mathit{x}+14$.
Since ab is the coefficient of ${\mathit{x}}^2$ on the left side of the equation and 15 is the coefficient of ${\mathit{x}}^2$ on the right side of the equation, it must be true that $\mathit{a}\mathit{b}=15$.Since $\mathit{a}+\mathit{b}=8$, it follows that $\mathit{b}=8-\mathit{a}$. Thus, $\mathit{a}\mathit{b}=15$ can be rewritten as $\mathit{a}\left(8-\mathit{a}\right)=15$, which in turn can be rewritten as ${\mathit{a}}^2-8\mathit{a}+15=0$. Factoring gives $\left(\mathit{a}-3\right)\left(\mathit{a}-5\right)=0$.
Thus, either $\mathit{a}=3$ and $\mathit{b}=5$, or $\mathit{a}=5$ and $\mathit{b}=3$. If $\mathit{a}=3$ and $\mathit{b}=5$, then Thus, one of the possible values of c is 31. If $\mathit{a}=5$ and $\mathit{b}=3$, then .
Thus, another possible value for c is 41. Therefore, the two possible values for c are 31 and 41.
Choice A is incorrect; the numbers 3 and 5 are possible values for a and b, but not possible values for c.
Choice B is incorrect;
if a = 5 and b = 3, then 6 and 35 are the coefficients of x when the expression $\left(5\mathit{x}+2\right)\left(3\mathit{x}+7\right)$ is expanded as $15{\mathit{x}}^2+35\mathit{x}+6\mathit{x}+14$.
However, when the coefficients of x are 6 and 35, the value of c is 41 and not 6 and 35. Choice C is incorrect;
if $\mathit{a}=3$ and $\mathit{b}=5$, then 10 and 21 are the coefficients of x when the expression $\left(3\mathit{x}+2\right)\left(5\mathit{x}+7\right)$ is expanded as $15{\mathit{x}}^2+21\mathit{x}+10\mathit{x}+14$.
However, when the coefficients of x are 10 and 21, the value of c is 31 and not 10 and 21.