SOLUTION: Please can you help me Find the values of a and b if ax^4+bx^3-8x^2+6x-6 has a remainder of 2x+1 when divided by x^2-1

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Question 1186024: Please can you help me
Find the values of a and b if ax^4+bx^3-8x^2+6x-6 has a remainder of 2x+1 when divided by x^2-1
Found 2 solutions by MathLover1, Edwin McCravy:
Answer by MathLover1(20850)   (Show Source): You can put this solution on YOUR website!


equal given factor to zero


=> or
substitute the value of the factor as in the expression and equate to zero plus reminder as is a factor of the expression.






....eq.1





....eq.2
from eq.1 and eq.2 we have




go to eq.1 and substitute
....eq.1


check
Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!

Here's another way to do it, by dividing directly with long division:

                    ax2+       bx+    (-8+a)
x2+0x-1)ax4+bx3-     8x2+      6x-         6
        ax4+0x3-     ax2
            bx3+(-8+a)x2+      6x
            bx3+     0x2-      bx
                (-8+a)x2+  (6+b)x-         6
                (-8+a)x2+      0x-    (-8+a)
                           (6+b)x+(-6+(-8+a))

For this remainder to be 2x+1,

6+b = 2        -6+(-8+a) = 1
  b = -4          -6-8+a = 1
                   -14+a = 1
                       a = 15
                       
Edwin
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