SATMathematics Practice Test 1

One can find the possible values of a and b in (ax+2)(bx+7) by using the given equation a + b = 8 and finding another equation that relates the variables a and b.
Since (ax+2)(bx+7)=15x2+cx+14, one can expand the left side of the equation to obtain abx2+7ax+2bx+14=15x2+cx+14.
Since ab is the coefficient of x2 on the left side of the equation and 15 is the coefficient of x2 on the right side of the equation, it must be true that ab=15.Since a+b=8, it follows that b=8−a. Thus, ab=15 can be rewritten as a(8−a)=15, which in turn can be rewritten as a2−8a+15=0. Factoring gives (a−3)(a−5)=0.
Thus, either a=3 and b=5, or a=5 and b=3. If a=3 and b=5, then Thus, one of the possible values of c is 31. If a=5 and 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 (5x+2)(3x+7) is expanded as 15x2+35x+6x+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 a=3 and b=5, then 10 and 21 are the coefficients of x when the expression (3x+2)(5x+7) is expanded as 15x2+21x+10x+14.
However, when the coefficients of x are 10 and 21, the value of c is 31 and not 10 and 21.