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Given P(x) = 4x3 + kx2 + 7x − 23 and the divisor is 2x-5.

Mathematics

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Answer

-8\text{-8}

Step-by-step solution

5 steps

  1. Apply the Remainder Theorem.

    The Remainder Theorem states that if a polynomial P(x)P(x) is divided by (ax−b)(ax-b), the remainder is P(ba)P\left(\frac{b}{a}\right).
    Given P(x)=4x3+kx2+7x−23P(x) = 4x^3 + kx^2 + 7x - 23 and the divisor is 2x−52x-5.
    Here, a=2a=2 and b=5b=5, so we need to evaluate P(52)P\left(\frac{5}{2}\right).
    The remainder is given as 7, so P(52)=7P\left(\frac{5}{2}\right) = 7.

  2. Substitute x=52x = \frac{5}{2} into the polynomial and set it equal to the remainder.

    4(52)3+k(52)2+7(52)−23=74\left(\frac{5}{2}\right)^3 + k\left(\frac{5}{2}\right)^2 + 7\left(\frac{5}{2}\right) - 23 = 7

  3. Simplify the terms.

    4(1258)+k(254)+352−23=74\left(\frac{125}{8}\right) + k\left(\frac{25}{4}\right) + \frac{35}{2} - 23 = 7 1252+25k4+352−23=7\frac{125}{2} + \frac{25k}{4} + \frac{35}{2} - 23 = 7

  4. Combine the constant terms.

    1252+352=1602=80\frac{125}{2} + \frac{35}{2} = \frac{160}{2} = 80

    So the equation becomes:

    80+25k4−23=780 + \frac{25k}{4} - 23 = 7 57+25k4=757 + \frac{25k}{4} = 7

  5. Isolate the term with kk and solve for kk.

    25k4=7−57\frac{25k}{4} = 7 - 57 25k4=−50\frac{25k}{4} = -50 25k=−50×425k = -50 \times 4 25k=−20025k = -200 k=−20025k = \frac{-200}{25} k=−8k = -8

    The value of the constant kk is -8\boxed{\text{-8}}.

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