Question

Find the value of p for which the polynomial 2x⁴+3x³+2(px²+3x+6 is divisible by (x+2)

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Answer 1

Answer :-

Value of p is - 1

Solution :-

Given

$\sf 2x^4 + 3x^3 + 2px^2 + 3x + 6 \: is \: divisible \: by \: (x + 2)$

It means it is a factor of (x + 2)

$\sf let \:f(x) = 2x^4 + 3x^3 + 2px^2 + 3x + 6$

First find the zero of (x + 2)

To find the zero equate x + 2 to 0

x + 2 = 0

x = -2

By factor theorem f(-2) = 0

$\sf f( - 2) = 0 \\ \\ \sf \rightarrow \:2( - 2)^4 + 3( - 2)^3 + 2p( - 2)^2 + 3( - 2) + 6 = 0 \\ \\ \sf \rightarrow 2(16) + 3( - 8) + 2p(4) - 6+ 6 = 0 \\ \\ \sf \rightarrow 32 - 24 + 8p = 0 \\ \\ \sf \rightarrow 8p = - 8 \\ \\ \sf \rightarrow p = - \dfrac{8}{8} \\ \\ \sf \rightarrow p = - 1$

$\Huge{\boxed{ \tt p = -1}}$

Answer 2

Answer:

Step-by-step explanation:

Given :-

2x⁴ + 3x³ + 2px² +3x + 6 divisible by (x+2)

To Find :-

The value of p

Solution :-

2x⁴ + 3x³ + 2px² + 3x

x + 2 = 0

x = -2

Put x = -2 in given polynomial.

p( -2) = 2(-2)⁴ + 3(-2)³ + 2p( -2)² + 3( -2) + 6

⇒ 32 - 24 + 8p - 6 + 6

⇒ 8 + 8p = 0

⇒ p = -8/8

⇒ p = -1

Hence, the value of p is -1.