Pearson Algebra 2 Common Core, 2011
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Exercise 4 Page 353

If a is a root of P(x)=0, then (x-a) is a factor of P(x).

P(x)=x^3-18/5x^2 + 19/5x -6/5

Practice makes perfect
We want to write a polynomial function with rational coefficients so that P(x)=0 has the given roots. 1, 2, and 35Recall that if a is a root of P(x)=0, then (x-a) is a factor of P(x). Let's simplify the polynomial by applying the Distributive Property. (x-1) * (x-2) * (x- 35) For simplicity, we will start by multiplying the first two factors and then the last factor to the product of the first two.
P(x)=(x-1)(x-2)(x-3/5)
P(x)=(x^2-2x-x+2)(x-3/5)
P(x)=(x^2-3x+2)(x-3/5)
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Simplify
P(x)=x^2(x-3/5)-3x(x-3/5)+2(x-3/5)
P(x)=x^3-3/5x^2-3x(x-3/5)+2(x-3/5)
P(x)=x^3-3/5x^2-3x^2+3(3/5)x+2(x-3/5)
P(x)=x^3-3/5x^2-3x^2+3(3/5)x+2x-2(3/5)
P(x)=x^3-3/5x^2-3x^2+9/5x+2x-6/5
P(x)=x^3-18/5x^2 + 19/5x -6/5
The polynomial with the zeros x=1, 2, and 35 is P(x)=x^3- 185x^2 + 195x - 65.