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STEP I 2002 question 5 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP I 2002 question 5 solution

f(0) = 0^{n} + a_{1} )^{n - 1} + ... + a_{n} = a_{n}

f(0) = (0 + k_{1})(0 + k_{2}) ... (0 + k_{n}) = k_{1} k_{2} ... k_{n}

Hence:

a_{n} = k_{1} k_{2} k_{3} ... k_{n}

f(1) = 1^{n} + a_{1} \left( 1^{n - 1} \right) + ... + a_{n} = 1 + a_{1} + a_{2} + ... + a_{n}

f(1) = (k_{1} + 1)(k_{2} + 1) ... (k_{n} + 1)

Hence:

1 + a_{1} + a_{2} + ... + a_{n} = (k_{1} + 1)(k_{2} + 1) ... (k_{n} + 1) .

f(-1) = (-1)^{n} + a_{1} (-1)^{n - 1} + ... + a_{n}

Hence, for even "n":

f(-1) = 1 - a_{1} + a_{2} - ... + a_{n}

For odd "n":

f(-1} = -1 + a_{1} - a_{2} + ... - a_{n}

But also:

f(-1) = (k_{1} - 1)(k_{2} - 1) ... (k_{n} - 1)

Hence, for even "n":

(k_{1} - 1)(k_{2} - 1) ... (k_{n} - 1) = 1 - a_{1} + a_{2} - ... + a_{n}

For odd "n":

(k_{1} - 1)(k_{2} - 1) ... (k_{n} - 1) = -1 + a_{1} - a_{2} + ... - a_{n} .

f(x) = x^{4} + 22x^{3} + 172x^{2} + 552x + 576

f(x) = 0 .

Notice that "n" is even.

Also notice that there is a solution of x = -2 , so k_{1} = 2 .

Hence:

a_{n} = 576 = 2k_{2} k_{3} k_{4}

1 + a_{1} + a_{2} + a_{3} + a_{4} = 1323 = (k_{1} + 1)(k_{2} + 1)(k_{3} + 1)(k_{4} + 1) = 3(k_{2} + 1)(k_{3} + 1)(k_{4} + 1)

Hence:

(k_{2} + 1)(k_{3} + 1)(k_{4} + 1) = 441 .

Also:

1 - a_{1} + a_{2} - a_{3} + a_{4} = 175 = (k_{1} - 1)(k_{2} - 1)(k_{3} - 1)(k_{4} - 1) = (k_{2} - 1)(k_{3} - 1)(k_{4} - 1) .

Now it is useful to consider the following:

(k_{2} + 1)(k_{3} + 1)(k_{4} + 1) = k_{2}k_{3}k_{4} + k_{2}k_{3} + k_{2}k_{4} + k_{3}k_{4} + k_{2} + k_{3} + k_{4} + 1 = 441

(k_{2} - 1)(k_{3} - 1)(k_{4} - 1) = k_{2}k_{3}k_{4} - k_{2}k_{3} - k_{2}k_{4} - k_{3}k_{4} + k_{2} + k_{3} + k_{4} - 1 = 175

Hence:

616 = 2 \left( k_{2}k_{3}k_{4} + k_{2} + k_{3} + k_{4} \right) .

The value of the product is known:

k_{2}k_{3}k_{4} = \frac{576}{2} = 288

Hence:

k_{2} + k_{3} + k_{4} = \frac{616}{2} - 288 = 308 - 288 = 20 .

Hence, the problem is to find three numbers which sum to twenty, and have a product of 288. Consider k_{2} = 8 , k_{3} = 6 , and k_{4} = 6 .

Hence, the roots of the equation are x = -2, \ or \ -6, \ or \ -6, \ or \ -8 .

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