Question

If (ax³+bx²-5x+2) has (x+2) as a factor and leaves a remainder 12 when divided by (x-2) , find the values of a and b.​

Answer 1

Answer:

Let f(x) = ax3 + bx2 - 5x + 2

Since x+2 is a factor of f(x), f(-2) = 0

So, -8a + 4b + 12 = 0

Since the remainder is 12 when f(x) is divided by x-2, f(2) = 12

So, 8a + 4b - 8 = 12

We have the system of equations: -8a + 4b = -12

8a + 4b = 20

Add the equations to obtain 8b = 8

b = 1

Since b = 1 and 8a + 4b = 20, 8a + 4 = 20

8a = 16

a = 2

Answer 2

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If (ax³+bx²-5x+2) has (x+2) as a factor and leaves a remainder 12 when divided by (x-2) , find the values of a and b.

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Let p(x) = ax³+ bx²- 5x+2,g(x)= x+2 and h(x) = x-2 .Then,

$g(x) = 0 ⟹ x + 2 = 0 = x ⟹ - 2$

$h(x) = 0 ⟹ x - 2 = 0 ⟹ x = 2.$

(x+2) is a factor of p(x) ⟹ p(-2) = 0.

$now \: p( - 2) = 0 ⟹ (a \times {( - 2)}^{3} + b \times {( - 2)}^{2} - 5 \times ( - 2) + 2) \\ ⟹ - 8a + 4b + 12 = 0 \\ ⟹ 8a - 4b = 12 ⟹ 2a - b = 3. \: \: \: \: \: \: ...(i)$

when p(x) is divided by (x-2), then the remainder is p(2).

$\therefore \: p(2) = 12 ⟹ ((a \times {2}^{3} ) + (b \times {2}^{2} ) - (5 \times 2) + 2) = 12 \\ ⟹ 8a + 4b = 20 ⟹ 2a + b = 5. \: \: \: \: \: \: ...(ii)$

On solving (i) and (ii) , we get a = 2 and b = 1.

Hence, a = 2 and b = 1.

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