Question

if the polynomials ×3+a×2+5and×3-2×2+a are divided by (×+2)leave the same remainder,find the value of a
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Answer 1

$\large \underline \bold{Question}$:-

if the polynomials (x³ + ax² + 5) and (x³ - 2x² + a) give the same remainder , when both are divided by (x + 2). then the value of a ?

$\large \underline \bold{Given}$:-

There are the two polynomials as -

1) $\sf{x^{3} + ax^{2} + 5}$

2) $\sf{x^{3} - 2x^{2} + a}$

both give the same remainder , when they are divided by (x + 2).

$\large \underline \bold{Solution}$:-

As we know ,

when both polynomials are divided by (x + 2) . remainder is same in each condition.

$\small \bold{(x^{3} + ax^{2} + 5) \: = \: (x^{3} - 2x^{2} + a) -(1)}$

So ,

$\qquad$(x + 2)$\:$=$\:$0

$\:$$\qquad \qquad$$\small \bold{x \: = \: - 2}$

$\small \bold{Now \: ,}$

On putting value of x in above eq. -

$\sf{(-2)^{3} + a(-2)^{2} + 5 \: = \: (-2)^{3} - 2(-2)^{2} + a}$

$\:$$\qquad$$\sf{-8 + a(4) + 5 \: = \: -8 - 2(4) + a}$

$\: \: \: \:$$\qquad$$\sf{-\cancel{8} + 4a + 5 \: = \: -\cancel{8} - 8 + a}$

$\: \: \: \: \: \: \:$$\qquad\qquad$$\sf{4a + 5 \: = \: - 8 + a}$

$\: \: \: \: \: \: \:$$\qquad\qquad$$\sf{4a - a \: = \: - 8 - 5}$

$\: \: \: \: \: \:$$\qquad\qquad\qquad$$\sf{3a \: = \: - 13}$

$\: \: \: \: \: \: \:$$\qquad\qquad\qquad$$\large \bold{a \: = \: -\dfrac{13}{3}}$

Answer 2

Answer:

$\huge\rm\green{\underline{\underline{Solution:–}}}$

given p(x)= x³+ax²+5

→ p(-2)= (-2)³+ a(-2)²+ 5

→ -8+ 4a+ 5

→ 4a- 3 → equation 1

→ q(x) = x³- 2x²+ a

→ q(-2) = (-2)- 2(-2)²+ a

→ -8- 8+ a

→ a- 16 → equation 2

$\rm\pink{Equate:–}$

Equation 1 and Equation 2

→ 4a- 3 = a- 16

→ 4a- a = 3- 16

→ 3a = -13

$\rm \: Answer→ a = \frac{ - 13}{3}$