Question

Find the value of a and b for which ax3-11x2+ax+b disuvible by x2-4x-5

Answer 1

$\textbf{\huge{ Hello Mate! }}$

Factors can be divided into sub factors

${x}^{2} - 4x - 5 = 0 \\ {x}^{2} - 5x + x - 5 = 0 \\ x(x - 5) + 1(x - 5) = 0 \\ (x + 1)(x - 5) = 0 \\ x = - 1 \: and \: 5$

Now, since they are factors so final value of p(x) will be 0.

$p(x) = a {x}^{3} - 11 {x}^{2} + ax + b \\ p( - 1) = a( { - 1)}^{3} - 11 {( - 1)}^{2} + a( - 1) + b \\ 0 = - a - 11 - a + b \\ 11 = - 2a + b \\ - 11 = 2a - b....(1) \\ \\ p(5) = a {(5)}^{3} - 11 {(5)}^{2} + a(5) + b \\ 0 = 125a - 275 + 5a + b \\ 275 = 130a + b....(2)$

On adding eq (1) and (2) we get

264 = 132a

2 = a

Subtituting value of a in eq (1)

-11 = 2(2) - b

- 11 - 4 = - b

15 = b

$\textsf{\red{ Hence value of a = 2 and b = 15 }}$

$\textbf{ Have great future ahead! }$

Answer 2

Answer:

x

2

−4x−5=0

x

2

−5x+x−5=0

x(x−5)+1(x−5)=0

(x+1)(x−5)=0

x=−1and5

Now, since they are factors so final value of p(x) will be 0.

\begin{gathered}p(x) = a {x}^{3} - 11 {x}^{2} + ax + b \\ p( - 1) = a( { - 1)}^{3} - 11 {( - 1)}^{2} + a( - 1) + b \\ 0 = - a - 11 - a + b \\ 11 = - 2a + b \\ - 11 = 2a - b....(1) \\ \\ p(5) = a {(5)}^{3} - 11 {(5)}^{2} + a(5) + b \\ 0 = 125a - 275 + 5a + b \\ 275 = 130a + b....(2)\end{gathered}

p(x)=ax

3

−11x

2

+ax+b

p(−1)=a(−1)

3

−11(−1)

2

+a(−1)+b

0=−a−11−a+b

11=−2a+b

−11=2a−b....(1)

p(5)=a(5)

3

−11(5)

2

+a(5)+b

0=125a−275+5a+b

275=130a+b....(2)

On adding eq (1) and (2) we get

264 = 132a

2 = a

Subtituting value of a in eq (1)

-11 = 2(2) - b

- 11 - 4 = - b

15 = b