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If x³+3x-4 is a factor of px³+qx+r and px³+qx+r leaves a remainder of 6 when it is divided by x-2, find the values of p,q,r

User Ben Sagmoe
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Final answer:

By equating coefficients and using the remainder theorem, we deduce that p=1, q=3, and r=-8 are the values that satisfy the given conditions for the polynomial px³+qx+r.

Step-by-step explanation:

If x³+3x-4 is a factor of px³+qx+r, then we can find values for p, q, and r by equating coefficients and using the remainder theorem. Since x³+3x-4 is a factor, the polynomial px³+qx+r can be written as (px³+qx+r) = (x³+3x-4)(ax⁴+bx²+cx+d). The remainder when px³+qx+r is divided by x-2 is 6, which implies p(2)³+q(2)+r = 6. Solving these expressions will give us the required values.

Let's find out the values by setting the coefficient of the highest power of x equal on both sides of the equation and substituting x=2 into the polynomial to use the remainder theorem. We conclude that:

  • p must be equal to 1 as it is the coefficient of x³ on both sides.
  • By substituting x=2 into px³+qx+r, we get p(2)³+q(2)+r = 6. With p as 1, this becomes 8+2q+r = 6, simplifying to 2q+r = -2.
  • Matching the linear coefficient, we get q = 3.
  • With q=3, we substitute back into 2q+r=-2 to find r. Thus 2(3)+r = -2, giving us r = -8.

Consequently, the values of p, q, and r are 1, 3, and -8 respectively.

User Agus Mathew
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