If $(a x+2)(b x+7)=15 x^{2}+c x+14$ for all values of $\mathrm{x}$, and $a+b=8$, what are the 2 possible values of $c$ ? 3 and 5 6 and 35 10 and 21 31 and 41
Solution
Step 1 ：The given equation is a quadratic equation. The coefficients of the quadratic terms on both sides of the equation must be equal. Similarly, the coefficients of the linear terms and the constant terms must also be equal.
Step 2 ：Given that \((a x+2)(b x+7)=15 x^{2}+c x+14\), we can expand the left side of the equation to get \(abx^2 + (7a + 2b)x + 14\).
Step 3 ：We know that \(ab = 15\) (from the coefficient of \(x^2\)) and \(a+b = 8\) (given in the question).
Step 4 ：The coefficient of \(x\) on the left side of the equation is \(7a + 2b\), which must be equal to \(c\).
Step 5 ：So, we need to find the possible values of \(c = 7a + 2b\).
Step 6 ：Solving the equations \(ab = 15\) and \(a + b = 8\), we get two possible solutions for \(a\) and \(b\), which are \((3, 5)\) and \((5, 3)\).
Step 7 ：Substituting these values into the expression for \(c\), we get two possible values for \(c\), which are 31 and 41.
Step 8 ：Final Answer: The two possible values for \(c\) are \(\boxed{31}\) and \(\boxed{41}\).
