$P_{1} (x) = 3 x^{2} + 10 x + 8$ and
$P_{2} (x) = x^{3} + x^{2} + 2 x + t$
are two polynomials.
When one of the factors of $P_{1} (x)$ divides $P_{2} (x)$ , 2 is the remainder obtained.

That factor is also a factor of the polynomial $P_{3} (x) = (x + 2) 2$ .

Find the value of ‘t’.

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Solution

The correct option is D
10

$3 x^{2} + 10 x + 8 = 3 x^{2} + 6 x + 4 x + 8$
$= 3 x (x + 2) + 4 (x + 2) = (x + 2) (3 x + 4)$

$P_{3} (x) = (x + 2) (2)$ .

Thus, the required factor is $(x + 2)$

When $P_{2} (x)$ is divided by $(x + 2)$ , the remainder = $P_{2} (- 2)$ .

$P_{2} (- 2) = (- 2)^{3} + (- 2)^{2} + 2 \times (- 2) + t$

$= - 8 + 4 - 4 + t = t - 8$

$P_{2} (2) = 2$

$⟹ t - 8 = 2$

$⟹ t = 10$ .

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