How do you find the polynomial function with roots 1, 7, and -3 of multiplicity 2?

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Answer 1 · 2015-08-10T20:25:21

$f (x) = 2 (x - 1) (x - 7) (x + 3) = 2 x^{3} - 5 x^{2} - 17 x + 21$ $f(x)=2(x-1)(x-7)(x+3)=2x^3-5x^2-17x+21$

If the roots are 1,7,-3 then in factored form the polynomial function will be:

$f (x) = A (x - 1) (x - 7) (x + 3)$ $f(x)=A(x-1)(x-7)(x+3)$

Repeat the roots to get the required multiplicity:

$f (x) = (x - 1) (x - 7) (x + 3) (x - 1) (x - 7) (x + 3)$ $f(x)=(x-1)(x-7)(x+3)(x-1)(x-7)(x+3)$

Answer 2 · 2015-08-11T21:15:39

The simplest polynomial with roots $1$ $1$ , $7$ $7$ and $- 3$ $-3$ , each with multiplicity $2$ $2$ is:

$f (x) = {(x - 1)}^{2} {(x - 7)}^{2} {(x + 3)}^{2}$ $f(x) = (x-1)^2(x-7)^2(x+3)^2$

$= x^{6} - 10 x^{5} - 9 x^{4} + 212 x^{3} + 79 x^{2} - 714 x + 441$ $=x^6-10x^5-9x^4+212x^3+79x^2-714x+441$

Any polynomial with these roots with at least these multiplicities will be a multiple of $f (x)$ $f(x)$ , where...

$f (x) = {(x - 1)}^{2} {(x - 7)}^{2} {(x + 3)}^{2}$ $f(x) = (x-1)^2(x-7)^2(x+3)^2$

$= {(x^{3} - 5 x^{2} - 17 x + 21)}^{2}$ $=(x^3-5x^2-17x+21)^2$

$= x^{6} - 10 x^{5} - 9 x^{4} + 212 x^{3} + 79 x^{2} - 714 x + 441$ $=x^6-10x^5-9x^4+212x^3+79x^2-714x+441$

...at least I think I've multiplied this correctly.

Let's check $f (2)$ $f(2)$ :

$2^{6} - 10 \cdot 2^{5} - 9 \cdot 2^{4} + 212 \cdot 2^{3} + 79 \cdot 2^{2} - 714 \cdot 2 + 441$ $2^6-10*2^5-9*2^4+212*2^3+79*2^2-714*2+441$

$= 64 - 320 - 144 + 1696 + 316 - 1428 + 441 = 625$ $=64-320-144+1696+316-1428+441=625$

${((2 - 1) (2 - 7) (2 + 3))}^{2} = {(1 \cdot - 5 \cdot 5)}^{2} = {(- 25)}^{2} = 625$ $((2-1)(2-7)(2+3))^2 = (1*-5*5)^2 = (-25)^2 = 625$