Question

Can u guys answer question 15

Answer 1

Answer:

$p(x) = x^{3} - ax^{2} + bx + 6$

ATQ

x - 2 is a factor of p(x)

thus ; x -2 = 0

x = 2

$p(2) = (2)^{3} - a(2)^{2} + b(2) + 6\\\\p(2) = 8 - 4a + 2b + 6\\\\p(2) = 14 - 4a + 2b$

now p(2) = 0

since x - 2 is a factor of p(x)

$14 - 4a + 2b= 0\\\\2b = 4a - 14\\\\2b = 2(2a - 7)\\\\b = 2a - 7$

also when p(x) is divided by x-3 it's remainder is 3

thus ; x - 3 = 0

x = 3

$p(3) = (3)^{3} - a(3)^{2} + b(3) + 6\\\\p(3) = 27 - 9a + 3b + 6\\\\p(3) = 33 - 9a + 3b$

now p(3) = 3

as given in the question

$33 - 9a + 3b = 3$

putting the value of b in the above equation

$33 - 9a + 2a - 7 = 3\\\\ - 7a + 26 = 3\\\\-7a = 3 - 26\\\\-7a = -23\\\\a = \frac{23}{7}$

now putting the value of a to find the value of b

$b = 2(\frac{23}{7}) - 7\\\\b = \frac{46 - 49}{7}\\\\b = \frac{-3}{7}$

I hope it's correct

Answer 2

${\huge {\overbrace {\underbrace{\blue{ANSWER: }}}}}$ 

Let p(x)=x^³+ax^²+bx+6

i ) it is given that , (x-2) is a factor of p(x) ,then

p(x)=0

=> 2^³+a(2)^²+b×2+6=0

=> 8+4a+2b+6=0

=> 4a+2b+14=0

Divide each term by 2 , we get

=> 2a+b+7=0 ---(1)

ii)It is given that, if p(x) divided by (x-3) leaves a remainder 3

=> p(3)=3

=> 3^³+a(3)^²+b×3+6=3

=> 27+9a+3b+6-3=0

=> 9a+3b+30=0

Divide each term by 3 , we get

=> 3a+b+10=0 ---(2)

Subtract equation (1) from equation (2) , we get

a + 3 = 0

=> a = -3

Now,

Substitute a=-3 in equation (1) , we get

2(-3)+b+7=0

=> -6+b+7=0

=> b+1=0

=> b = -1

Therefore,

Value of a = -3

value of b =-1