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A) Let $$f(x)$$ be a polynomial of degree $$4$$ with rational coefficients which has $$1+2\sqrt{3}$$ and $$3-\sqrt{2}$$ as roots, and such that $$f(0) = -154$$ Find $$f(1)$$

B) Post your second question on a second post.

May 6, 2019
edited by Melody  May 6, 2019

#1
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The roots are  (1 + √12) ( 1-√12) (3 - √2)  and (3 + √2)

So........the polynomial is

a [ x - (1+√12) ] [ x - (1 - √12) ] [ x - ( 3 - √2) ] [ x- (3 + √2)]

Simplifying....we have that......

a [x^4 - 8 x^3 + 8 x^2 + 52 x - 77]  =

And since f(0)  = -154

Then  -77a  = -154

So....a  = 2

So....the polynomial is

2x^4 - 16x^3 + 16x^2 + 104x - 154

And f(1)  =   2 - 16 + 16 + 104 - 154  =  -48   May 7, 2019
#2
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A)

Let $$f(x)$$ be a polynomial of degree $$4$$  with rational coefficients which has $$1+2\sqrt{3}$$ and $$3-\sqrt{2}$$
as roots, and such that $$f(0) = -154$$.
Find $$f(1)$$
.

Vieta:

$$\begin{array}{|rcll|} \hline \mathbf{f(x)} &=& \mathbf{ax^4+bx^3+cx^2+dx+e} \quad | \quad \text{polynomial of degree 4} \\\\ \dfrac{e}{a} &=& x_1x_2x_3x_4 \quad | \quad \text{Vieta-theorem},\ f(0) =e \\\\ \dfrac{f(0)}{a} &=& x_1x_2x_3x_4 \\\\ \mathbf{a} &=& \mathbf{\dfrac{f(0)}{x_1x_2x_3x_4}} \\\\ && \boxed{ f(0)=-154\\ x_1 = 1+2\sqrt{3}\\ x_2 = 1-2\sqrt{3}\\ x_3 = 3-\sqrt{2}\\ x_4 = 3+\sqrt{2} } \\\\ a &=& \dfrac{-154}{(1+2\sqrt{3})(1-2\sqrt{3})(3-\sqrt{2})(3+\sqrt{2})} \\\\ &=& \dfrac{-154}{(1-4\cdot 3)(9-2)} \\\\ &=& \dfrac{-154}{(-11)\cdot 7} \\\\ &=& \dfrac{ 154}{77} \\\\ \mathbf{a} &=& \mathbf{2} \\ \hline \end{array}$$

$$\mathbf{f(1)=\ ?}$$

$$\begin{array}{|rcll|} \hline \mathbf{f(x)} &=& \mathbf{ a(x-x_1)(x-x_2)(x-x_3)(x-x_4) } \\\\ f(1) &=& 2\left(1-(1+2\sqrt{3})\right) \left(1-(1-2\sqrt{3})\right) \left(1-(3-\sqrt{2})\right) \left(1-(3+\sqrt{2})\right) \\ &=& 2(2\sqrt{3}) (-2\sqrt{3}) (-2+\sqrt{2}) (-2-\sqrt{2})) \\ &=& 2(2\sqrt{3}) (-2\sqrt{3}) \left((-2)^2-2\right) \\ &=& 2(2\sqrt{3}) (-2\sqrt{3}) \cdot 2\\ &=& -16\cdot 3 \\ \mathbf{f(1)} &=& \mathbf{-48} \\ \hline \end{array}$$ May 7, 2019