Question

If ax² + bx + c is divided by x + 3, x – 5, and x – 1, the remainders are 21, 61 and 9 respectively. Find a, b and c. (Use Gaussian elimination method.)

Answer 1

P(x) = ax² + bx + c.

When P(x) is divided by x + 3, x – 5 and x – 1.

The remainders are respectively P(– 3), P(5) and P(1).

We are given that P(– 3) = 21

P(5) = 61

P(1) = 9

Now P(– 3) = 21

⇒ a(– 3)² + b(– 3) + c = 21

⇒ 9a – 3b + c = 21  ........(1)

P(5) = 61

⇒ a(5)² + b(5) + c = 61

⇒ 25a + 5b + c = 61  .......(2)

P(1) = 9

⇒ a(1)² + b(1) + c = 9

⇒ a + b + c = 9   .......(3)

Now the matrix form of the above three equations is

${\displaystyle \left[\begin{array}{ccc}9& -3& 1\\ 25& 5& 1\\ 1& 1& 1\end{array}\right]\left[\begin{array}{c}\text{a}\\ \text{b}\\ \text{C}\end{array}\right]=\left[\begin{array}{c}21\\ 61\\ 9\end{array}\right]}$

(i.e) AX = B

The augmented matrix (A, B) is

[A, B] = ${\displaystyle \left[\begin{array}{cccc}9& -3& 1& 21\\ 25& 5& 1& 61\\ 1& 1& 1& 9\end{array}\right]}$

${\displaystyle ˜\left[\begin{array}{cccc}1& 1& 1& 9\\ 25& 5& 1& 61\\ 9& -3& 1& 21\end{array}\right]{\text{R}}_1\leftrightarrow {\text{R}}_3}$

${\displaystyle \left[\begin{array}{cccc}1& 1& 1& 9\\ 25& 5& 1& 61\\ 9& -3& 1& 21\end{array}\right]˜\left[\begin{array}{cccc}1& 1& 1& 9\\ 0& -20& -24& -164\\ 0& -12& -8& -60\end{array}\right]\begin{array}{l}{\text{R}}_2\to {\text{R}}_2-25{\text{R}}_1\\ {\text{R}}_3\to {\text{R}}_3-9{\text{R}}_1\end{array}}$

${\displaystyle ˜\left[\begin{array}{cccc}1& 1& 1& 9\\ 0& -20& -24& -164\\ 0& 12& 8& 60\end{array}\right]{\text{R}}_3\to -{\text{R}}_3}$

${\displaystyle ˜\left[\begin{array}{cccc}1& 1& 1& 9\\ 0& -20& -24& -164\\ 0& 0& -32& -192\end{array}\right]{\text{R}}_3\to -5{\text{R}}_3+3{\text{R}}_2}$

The above matrix is in echelon form now writing the equivalent equations.

${\displaystyle \left[\begin{array}{ccc}1& 1& 1\\ 0& -20& -24\\ 0& 0& -32\end{array}\right]\left[\begin{array}{c}\text{a}\\ \text{b}\\ \text{c}\end{array}\right]=\left[\begin{array}{c}9\\ -164\\ -192\end{array}\right]}$

(i.e) a + b + c = 9

– 20b – 24c = – 164

– 32c = – 192

From (3)

⇒ c = ${\displaystyle \frac{-192}{-32}}$ = 6

Substituting c = 6 in (2) we get

– 20b – 24(6) = – 164

⇒ – 20b = – 164 + 144 = – 20

⇒ b = 1

Substituting c = 6, b = 1 in (1) we get

a + 1 + 6 = 9

⇒ a = 9 – 7 = 2

So a = 2, b = 1, c = 6