Question

If 2x3 + ax2 + bx - 2 has a factor of (x+2) and leaves a remainder 7 when divided by 2x-3 .Find the values of a & b and Hence factories completely.

Tell me It's urgent..​

Answer 1

Correct Question

If 2x³ + ax² + bx - 2 has a factor of (x+2) and leaves a remainders 0 and 7 when divided by (2x-3)

Find out

Value of a and b

Solution

★ Given polynomial

• 2x³ + ax² + bx - 2

✞ (x + 2) is a factor of given polynomial

➞ x + 2 = 0

➞ x = - 2

• Remainder = 0
• 2x³ + ax² + bx - 2 = 0

✞ Put the value of x

➞ 2(-2)³ + a(-2)² + b(-2) - 2 = 0

➞ 2 × (-8) + 4a - 2b - 2 = 0

➞ -16 + 4a - 2b - 2 = 0

➞ 4a - 2b = 18

➞ 2(2a - b) = 18

➞ 2a - b = 9 ----(i)

$\rule{200}3$

✞ (2x - 3) is a factor of given polynomial

➞ 2x - 3 = 0

➞ x = 3/2

✞ Put the value of x

• Remainder = 7
• 2x³ + ax² + bx - 2 = 7

➞ 2(3/2)³ + a(3/2)² + 3b/2 - 2 = 7

➞ 2 × 81/8 + 9a/4 + 3b/2 = 7 + 2

➞ 27/4 + 9a/4 + 3b/2 = 9

➞ 27 + 9a + 6b/4 = 9

➞ 27 + 9a + 6b = 9 × 4

➞ 9a + 6b = 36 - 27

➞ 3(3a + 2b) = 9

➞ 3a + 2b = 3 ----(ii)

$\rule{200}3$

★ Multiply (i) by 2 and (ii) by 1

• 4a - 2b = 18
• 3a + 2b = 3

★ Add both the equations

➞ 4a - 2b + 3a + 2b = 18 + 3

➞ 7a = 21

➞ a = 21/7 = 3

★ Putting the value of a in eqⁿ (ii)

➞ 3a + 2b = 3

➞ 3 × 3 + 2b = 3

➞ 9 + 2b = 3

➞ 2b = 3 - 9

➞ 2b = - 6

➞ b = -6/2 = - 3

Hence,

• Required value of a = 3
• Required value of b = - 3

$\rule{200}3$

Answer 2

◘ Correct question :-

2x³ + ax² + bx - 2 leaves remainder 7 and 0 when when divided by (2x - 3) and (x + 2) respectively. Find the values of a & b and hence factories completely.

✱ Given ✱

• When 2x³ + ax² + bx - 2 is divided by (2x - 3), the remainder is 7.
• When 2x³ + ax² + bx - 2 is divided by (x + 2), the remainder is 0.

✱ To Find ✱

• The value of a and b.

✱ Solution ✱

Let  2x³ + ax² + bx - 2 = p(x).

Zero of (2x - 3) :-

2x - 3 = 0

⇒ 2x = 3

⇒ x = 3/2

A/q,

$\sf{p\bigg(\dfrac{3}{2}\bigg)=2 \bigg( \dfrac{3}{2}\bigg)^3+a \bigg( \dfrac{3}{2} \bigg)^2 + b\bigg( \dfrac{3}{2}\bigg)-2=7}\\\\\displaystyle{\sf{\longrightarrow 2 \times \frac{27}{8}+a \times \frac{9}{4} +\frac{3b}{2}-2=7 }}\\\\\displaystyle{\sf{\longrightarrow \frac{27}{4}+\frac{9a}{4}+\frac{3b}{2}-2=7 }}\\\\\displaystyle{\sf{\longrightarrow \frac{27+9a+6b-8}{4} =7}}\\\\\displaystyle{\sf{\longrightarrow \frac{19+9a+6b}{4} =7}}\\\\\displaystyle{\sf{\longrightarrow 19+9a+6b=28}}$

$\displaystyle{\sf{\longrightarrow 9a+6b=28-19}}\\\\\displaystyle{\sf{\longrightarrow 9a+6b=9}}\\\\\displaystyle{\sf{\longrightarrow 3a+2b=3}}\:\:\:\: \cdots\sf{(i)}$

$\rule{180}2$

Zero of (x + 2) :-

x + 2 = 0

⇒ x = -2

Again, A/q,

$\displaystyle{\sf{p(-2)=2(-2)^3+a(-2)^2+b(-2)-2=0 }}\\\\\displaystyle{\sf{\longrightarrow 2\times-8+4a-2b-2=0}}\\\\\displaystyle{\sf{\longrightarrow -16+4a-2b-2=0}}\\\\\displaystyle{\sf{\longrightarrow 4a-2b-18=0}}\\\\\displaystyle{\sf{\longrightarrow 4a-2b=18}}\\\\\displaystyle{\sf{\longrightarrow 2a-b=9}}\:\:\:\: \cdots \sf{(ii)}$

$\rule{180}2$

Solving eq. (i) and eq. (ii) :-

From eq. (ii) -

2a = 9 + b

⇒ b = 2a - 9

→ Putting the value of b in eq. (i) :-

3a + 2b = 3

⇒ 3a + 2(2a - 9) = 3

⇒ 3a + 4a - 18 = 3

⇒ 7a = 3 + 18

⇒ 7a = 21

⇒ a = 21/7

⇒ a = 3

Putting the value of a :-

b = 2a - 9

⇒ b = 2(3) - 9

⇒ b = 6 - 9

⇒ b = -3