# What is m if p=2x^4-mx^3+x^2-7 divided by x+2 to have the reminder r=4 ?

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### 2 Answers

P =2x^4-mx^3+x^2-7 divided by x+2.

Let P = P(x) =2x^4-mx^3+x^2-7.

Since P(x) divided by (x+2) gives a remainder r=4, we write b division lgorithm,

P(x) /(x+2 = Q(x) + r the remainder. , where Q(x) is a thirs degree expression. Now multiply both soides by (x+2).

P(x) = (x+2)Q(x) + 4, as r = 4.

2x^4-mx^3+x^2-7 = (x+2)Q(x) +4.

Put x= -2 on both sides>

2(-2)^4 -m(-2)^3 +(-2)^2 -7 = 0*Q(-2) +4

32+8m +4-7 = 0+4

8m -29= 4

8m = 4-29 = -25.

m = -25/8 .

Therefore the value of m = -25/8.

According to the fundamental theorem of algebra, if P(x) is divided by (x+2) and the reminder is 4, then we could write:

P(-2)=4 (1)

We'll apply the rule of division with reminder:

2X^4 - mX^3 + X^2 - 7= Q(x+2) + 4

But P(-2)=4, so, we'll substitute x by -2 in the expression of polynomial P(x).

P(-2) = 2(-2)^4 - m(-2)^3 + (-2)^2 - 7

P(-2) = 32 + 8m + 4 - 7

We'll substitute P(-2) by 4:

4 = 8m + 29

We'll use the symmetric property:

8m + 29 = 4

We'll subtract 29:

8m = 4 - 29

8m = -25

We'll divide by 8:

**m = -25/8**

**The polynomial P(x), whose reminder is 4 when it's divided by (x+2), is determined for m = -25/8.**