 # Week 21

Example 1. Find the value of 23 × 15.

Example 2. The function f satisfies f(2 + x) = f(2 − x) for all real numbers x. Moreover, f(x) = 0 has exactly four distinct real roots. What is the sum of these roots?

Of course, you can find the solution just below, but it is highly recommended that, you first try to solve it on your own.

Just remember the words of Paul Halmos, who says “the only way to learn mathematics is to do mathematics”.

Solution 1. Put the data in a calculator to get 23 × 15 = 345.

Alternatively, a direct multiplication yields

2 3

×  1 5

1 1 5

2 3 ×

3 4 5

As another alternative, left distributive property gives

23 × 15 = (20 + 3) × 15

= 20 × 15 + 3 × 15

= 300 + 45

= 345

Alternatively, applying right distributive property gives

23 × 15 = 23 × (10 + 5)

= 23 × 10 + 23 × 5

= 230 + 115

= 345

As another alternative, using identities

23 × 15 = (20 + 3) × (10 + 5)

= 20 × 10 + 20 × 5 + 3 × 10 + 3 × 5

= 200 + 100 + 30 + 15

= 345

Finally, using identities

23 × 15 = (19 + 4) × (19 − 4)

= 192 – 42

= 361 – 16

= 345

Solution 2. Suppose that r is a root of f(x) = 0. Due to the fact f(2 + x) = f(2 − x) for all x, we have

0 = f(r)

= f[2 + (r − 2)]

= f[2 − (r − 2)]

= f(4 − r),

so 4 − r is also a root. Moreover, the sum of this pair is r + (4 − r) = 4.

Since there are two pairs of roots of this type, the sum of the four roots is 8.

An elementary function satisfying the property given in the statement of the exercises is f(x) = x(x − 1)(x − 3)(x − 4).