Question

If p(x) =2x4 + 3x3 3x2 2x + 5 is divided by 2x2 + 3x 1, then the remainder is x a. Find a.

Answer 1

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If p(x)=2x^4+3x^3-3x^2-2x+5 is divided by 2x^2+3x-1 then the remainder is x-a.find a.

P(x) = 2x⁴ + 3x³ + 3x² + 2x + 5

Rearranging and rewriting the above equation so as to get the divisor 2x² + 3x + 1,

P(x) = 2x⁴ + 3x³ + x² + 2x² + 2x +1+4

= P(x) = x²(2x² + 3x + 1) + 2x² + 2x + 1 + 4

Again for making the divisor adding and subtracting x in above equation

P(x) = x²(2x² + 3x + 1) + 2x² + 2x + x + 1 - x + 4

= P(x) = x²(2x² + 3x + 1) + (2x² + 3x + 1) - x + 4

Now as the first two terms of P(x) contain the divisor they are divisible by it leaving remainder zero.Hence the remainder is equal to the last term .

= Remainder = -x + 4

Thus a = 4.
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Answer 2

The value of a is 4.

GIVEN

Dividend -

$p(x) = {2x}^{4} + {3x}^{3} + {3x}^{2} + 2x + 5$

Divisor -

${2x}^{2} + 3x + 1$

Remainder-

x + a

TO FIND

The value of a.

SOLUTION

We can simply solve the above problem as follows-

We first rearrange the dividend expression to get the given divisor,

$p(x) = {2x}^{4} + {3x}^{3} + {3x}^{2} + 2x + 5$

$= p(x) = {2x}^{4} + {3x}^{3} + {x }^{2} + {2x}^{2} + 2x + 1 + 4$

Simplifying further -

$p(x) = {x}^{2} ({2x}^{2} + {3x} + 1 ) + {2x}^{2} + 2x + 1 + 4$

Adding and subtracting x in the above equation ;

$p(x) = {x}^{2} ({2x}^{2} + {3x} + 1 ) + {2x}^{2} + 2x +x + 1 - x + 4$

We can also write it as -

$p(x) = {x}^{2} ({2x}^{2} + {3x} + 1 ) + ({2x}^{2} + 2x +x + 1) - x + 4$

as the first two terms contain the divisor so they exactly divisible by the divisor, the above equation will leave '-x+4' as remainder.

So,

-x+ 4 = x + a

a = 4

Hence, The value of a is 4.

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