Question

1.
Find the values of a and b so that the polynomial x^3 + 10x^2+ ax+ b
is exactly divisible by (x - 1) as well as (x - 2)​

Answer 1

$\huge\mathbb\blue{ANSWER!!!!}$

FOR FINDING A AND B PUT VALUE OF X FIRST . GIVEN (X-1) (X-2) PUT P (X)=0

x=1,2

put it in given equation

${x}^{3} - {10x}^{2} + ax + b$

$\huge\boxed{\sf{TAKE \: X=1 }}$

$0 = {1}^{3} - {10 \times 1}^{2} + a \times 1 + b \\ \\ 0 = 1 - 10 + a + b \\ \\ 0 = - 9 + a + b - - - (1)$

$\huge\boxed{\sf{TAKE \: \: X = 2 }}$

$0 = {2}^{3} - {10 \times 2}^{2} + a \times 2 + b \\ \\ 0 = 8 - 40 + 2a + b \\ \\ 0 = - 32 + 2a + b - - - (2)$

SUBTRACT 1eq from 2 eq

$= > - 9 + a + b - ( - 32+ 2a + b) \\ \\ = > 23 - a = 0 \\ \\ = > a = 23$

putting A in 1eq

$= > 0 = - 9 + 23 + b \\ \\ b = - 14$

$\huge\boxed{\sf{A=23 , B= -14}}$

Answer 2

Polynomial , x³ +10x² + ax + b is exactly divisible by ( x -1) and (x -2)

it means x = 1 and 2 are the zeors of given polynomial .

so,
put x = 1

(1)³ + 10(1)² +a(1) + b = 0

1 + 10 + a + b = 0

a + b = -11 --------(1)

again put x =2

(2)³ +10(2)² +a(2) + b =0

8 + 40 + 2a + b = 0

2a + b = -48 ----------(2)

solve equations (1) and (2)

a = -37

b = 26