Question

Find the values of a and b if (x -2) is a factor of f(x) = x3 + ax2 + bx +26 and
f(3) = 32
please solve​

Answer 1

Answer:

a=7 , b=-31

Step-by-step explanation:

given (x-2) is a factor of x³+ax²+bx+26

i.e 2 is a zero of the polynomial

0=2³+a2²+b2+26

0=8+4a+2b+26

0=4+2a+b+13

2a+b+17=0....(1)

also 32=3³+a3²+b3+26 (given)

6=27+9a+3b

9a+3b+21=0

3a+b+7=0....(2)

subtracting (1) and (2) we get

-a+7=0

a=7

put a=7 in equation (1) we get

14+b+17=0

b= -31

hope it helped.

Answer 2

Given :-

• ( x -2 ) is a factor of f(x) = x³ + ax² + bx + 26.
• The value of f(3) = 32 .

To Find :-

• The value of a and b .

Solution :-

Given that ( x-2) is a factor of x³ + ax² + bx + 26.

The value of f(3) = 32 . So , if now x - 2 is a factor of f(x) , then x = 2 will be a zero of f(x) .

$=> f(x) = x^3 +ax^2+bx + 26 \\ => f(2) = 2^3 + 2^2.a + 2b + 26 = 0 \\ => 8 + 4a + 2b + 26 = 0 \\ =>4a + 2b + 34 = 0 \\=> 2 ( 2a + b + 17 ) = 0 \\=> 2a + b + 17 = 0$ ...........(i)

According to second info :-

Value of f(3) = 32

$=> f(3)=32\\=> 3^3+3^2.a+3b+26 = 32 \\=> 27 + 9a + 3b = 32 - 26 \\=> 9a + 3b = 6 - 27 \\=> 9a + 3b = (-21)\\ => 3 ( 3a + b ) = (-21)\\=> 3a + b = -7 \\=> 3a+b+7 = 0$ . ................(ii)

Subtract (ii) and (i) :-

$=> (3a + b +7)-(2a+b+17)=0-0 \\=> 3a + b + 7 - 2a - b -17 = 0 \\=> a - 10 = 0 \\\bf =>a = 10$

Put this is (i)

$=> 2a+b+17 = 0 \\=> 2(10)+b+17 = 0\\=> 20+17+b = 0 \\=> b + 37 = 0 \\\bf=> b = (-37 )$

$\Large\boxed{\pink{\sf Value\:of\:a=10\:\&\:b\:=\:(-37)}}$