sat suite question viewer

Advanced Math / Equivalent expressions Difficulty: Hard

One of the factors of $2 x^{3} + 42 x^{2} + 208 x$ is $x + b$ , where $b$ is a positive constant. What is the smallest possible value of $b$ ?

Back question 50 of 102 Next

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102

Explanation

The correct answer is $8$ . Since each term of the given expression, $2 x^{3} + 42 x^{2} + 208 x$ , has a factor of $2 x$ , the expression can be rewritten as $2 x (x^{2}) + 2 x (21 x) + 2 x (104)$ , or $2 x (x^{2} + 21 x + 104)$ . Since the values $8$ and $13$ have a sum of $21$ and a product of $104$ , the expression $x^{2} + 21 x + 104$ can be factored as $(x + 8) (x + 13)$ . Therefore, the given expression can be factored as $2 x (x + 8) (x + 13)$ . It follows that the factors of the given expression are $2$ , $x$ , $x plus 8$ , and $x plus 13$ . Of these factors, only $x plus 8$ and $x plus 13$ are of the form $x + b$ , where $b$ is a positive constant. Therefore, the possible values of $b$ are $8$ and $13$ . Thus, the smallest possible value of $b$ is $8$ .