Question

When a polynomial 2x^3+3x^2+ax+b is divided by (x-2) leaves remainder 2, and (x+2) leaves remainder -2.Find a and b.​

Answer 1

Answer:-

Given:

p(x) = 2x³ + 3x² + ax + b when divided by x - 2 and x + 2 leaves the remainders 2 , - 2.

So,

✯ g(x) = x - 2

⟹ 0 = x - 2 [ g(x) = 0 ]

⟹ x = 2

★ p(2) = 2(2)³ + 3(2)² + a(2) + b

⟹ - 2 = 2 * 8 + 3 * 4 + 2a + b

⟹ - 2 - 16 - 12 - 2a = b

⟹ - 30 - 2a = b -- equation (1)

Similarly,

✯ g(x) = x + 2

⟹ 0 = x + 2

⟹ x = - 2

★ p( - 2) = 2( - 2)³ + 3( - 2)² + a ( - 2) + b

Substitute the value of b from equation (1).

⟹ 2 = 2 * ( - 8) + 3 * 4 - 2a - 30 - 2a

⟹ 2a + 2a = - 16 + 12 - 2 - 30

⟹ 4a = - 24

⟹ a = - 24/4

⟹ a = - 6

Substitute the value of a in equation (1).

⟹ - 30 - 2( - 6) = b

⟹ - 30 + 12 = b

⟹ - 18 = b

Therefore,

• a = - 6
• b = - 18.

Answer 2

Answer:

$\huge\tt{\boxed{\underbrace{\overbrace{\bold\color{blue}{☛AnSwEr☚}}}}}$

$\green{ \underline{ \boxed{ \sf{Given:-}}}}$

• p(x) = 2x³ + 3x² + ax + b when divided by x - 2 and x+2 leaves remainder 2,-2

$\green{ \underline{ \boxed{ \sf{To\:find:-}}}}$

• a and b

$\huge\underbrace\mathfrak\color{lime}{ †\: Solution:–}$

• g(x)=x-2

=>0=x-2[g(x)=0]

=>x=2

• p(2)=2(2)³ +3(2)² +a(2)+b

=>-2=2 ×8+3×4+2a+b

=> -2-16-12-2a=b

=>-30-2a=b-e equation (1)

Similarly,

• g(x)=x+2

=>0=x+2

=> x=-2

• p(-2)=2(-2)^{3}+3(-2)^{2}+a(-2)+b

Substitute the value of b from equation (1).

=>2=2^{*}(-8)+3^{*}4-2a-30-2a

=>2a+2a=-16+12-2-30

=> 4a=-24

=> a=-24/4

=> a=-6

Substitute the value of a in equation (1).

=>-30-2(-6)=b

=>-30+12=b

=>-18=b

$\underbrace\mathfrak\color{aqua}{ †\: Therefore, }$

• a = - 6
• b = - 18 .