It is given that f(x)=2x3+ax2+bx−10f(x)=2x^3+ax^2+bx-10, where aa and bb are constants, has a factor of (x−1)2(x-1)^2.

  1. (a)

    Find the value of aa and of bb.

  2. (b)

    It is given that g(x)=f(x)+kg(x)=f(x)+k, where kk is a constant. Using the values of aa and bb found in (a), find the value of kk such that g(x)g(x) is divisible by 2x−12x-1.

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Worked solution

7(b)

1. State the values of $a$ and $b$ from part (a) and write the expression for $g(x)$

From part (a), a=−14a = -14 and b=22b = 22.
f(x)=2x3−14x2+22x−10f(x) = 2x^3 - 14x^2 + 22x - 10
g(x)=2x3−14x2+22x−10+kg(x) = 2x^3 - 14x^2 + 22x - 10 + k

2. Apply the Factor Theorem to find $k$

Since g(x)g(x) is divisible by 2x−12x - 1, by the Factor Theorem, g(12)=0g\left(\frac{1}{2}\right) = 0.
2(12)3−14(12)2+22(12)−10+k=02\left(\frac{1}{2}\right)^3 - 14\left(\frac{1}{2}\right)^2 + 22\left(\frac{1}{2}\right) - 10 + k = 0
2(18)−14(14)+11−10+k=02\left(\frac{1}{8}\right) - 14\left(\frac{1}{4}\right) + 11 - 10 + k = 0
14−72+1+k=0\frac{1}{4} - \frac{7}{2} + 1 + k = 0
0.25−3.5+1+k=00.25 - 3.5 + 1 + k = 0
−2.25+k=0-2.25 + k = 0
k=2.25k = 2.25

Answer

k=2.25k = 2.25

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