Jose asked in Science & MathematicsMathematics · 1 month ago

# How do I solve this?

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• 1 month ago

When a polynomial p(x) is divided by the binomial (x - a), the remainder is p(a).

So, with p(x) = x^4 + ax^3 + bx + c, the given remainders mean that:

p(1) = 14

p(-1) = 0

p(-2) = -16

Evaluate p for those values of x on the left:

1 + a + b + c = 14

1 - a - b + c = 0

16 - 8a - 2b + c = -16

Three linear equations in three unknowns.  Add the first two and both b and c cancel out:

2 + 2c = 14 = 0

2c = 12

c = 6

Subtract them instead and get one reduced equation:

2a + 2b = 14; a + b = 7

Plug c=6 into the 3rd equation to get another reduced equation:

16 - 8a - 2b + 6 = -16

16 + 16 + 6 = 8a + 2b

8a + 2b = 38

4a + b = 19

Subtract a + b = 7 and get

3a = 19 - 7 = 12

a = 4

...and b = 7 - a = 3

I just typed all that on the run to show the math.  You should carry out those steps yourself in case I made a typo along the way.

• 1 month ago

Ruffini

(x⁴+ax³+bx+c)/(x+1)=x³+(a-1)x²-(a-1)x+(a-1+b) with remaind (1-a-b+c)=0

(x⁴+ax³+bx+c)/(x-1)=x³+(a+1)x²+(a+1)x+(a+1+b) with remaind (1+a+b+c)=14

(x⁴+ax³+bx+c)/(x+2)=x³+(a-2)x²+(4-2a)x+(4a-8+b) with remaind (16-8a-2b+c)=-16

Now

-a-b+c=-1

a+b+c=13

-8a-2b+c=-32

Solve

a=4

b=3

c=6

Try

(x⁴+4x³+3x+6)/(x+1)=(x³+3x²-3x+6)(x+1)+0