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

Question

Jocelynn Vega · Accepted Answer

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)

To obtain the desired multiplicity, repeat the roots:

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

Stephanie Oconnell · Accepted Answer

Any polynomial with these roots with at least these multiplicities will be a multiple of

f (x)

, where...

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

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

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

At least I believe I correctly multiplied this.
Let's check

f (2)

:

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

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

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

Answered question