Question

Pi (x) = 3x2 + 10x + 8 and
P(x) = 1 + x2 + 2x +t
are two polynomials.
When one of the factors of Pi (3) divides P2 (x), 2 is the
remainder obtained
That factor is also a factor of the polynomial P3(x) = 2(x + 2)
Find the value of 't​

Answer 1

Question :- P1(x) = 3x² + 10x + 8 and P2(x) = x³ + x² + 2x + t

are two polynomials. When one of the factors of P1(x) divides P2(x), 2 is the remainder obtained. That factor is also a factor of the polynomial P3(x) = 2(x + 2). Find the value of 't'. ?

Solution :-

Solving P1(x) first,

→ 3x² + 10x + 8

→ 3x² + 6x + 4x + 8

→ 3x(x + 2) + 4(x + 2)

→ (3x + 4)(x + 2)

Putting both equal to 0, we get,

→ 3x + 4 = 0

→ 3x = (-4)

→ x = (-4/3) .

Or,

→ x + 2 = 0

→ x = (-2).

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Now, Solving P3(x),

→ 2(x + 2) = 0

→ x + 2 = 0

→ x = (-2).

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Given that, P3(x) and P1(x) has a common factor.

Therefore, that factor is Equal to (-2).

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Now, we have given that, when this divides P2(x), 2 is the remainder obtained. .

So, By remainder theorem , we get,

→ P2(x) = x³ + x² + 2x + t

→ P(-2) = x³ + x² + 2x + t = 2

→ (-2)³ + (-2)² + 2*(-2) + t = 2

→ (-8) + 4 - 2 + t = 2

→ t - 6 = 2

→ t = 2 + 6

→ t = 8. (Ans.)

Hence, the value of t is 8.