Build a polynomial backwards
Knowing the zeroes—or just their sum and product—is enough to construct a quadratic polynomial.
Knowing the zeroes—or just their sum and product—is enough to construct a quadratic polynomial.
(x-alpha)(x-beta)=x^2-(alpha+beta)x+alpha beta. This expansion explains the minus sign before the sum.
If the sum is S and product is P, a quadratic polynomial is x^2-Sx+P. Substitute S with its sign: when S=-4, the middle term is -(-4)x=+4x.
If p(k)=0, then lambda p(k)=0 for every non-zero lambda. Therefore x^2-Sx+P is the simplest monic choice, while any non-zero scalar multiple has the same zeroes.
Form a quadratic polynomial whose zeroes have sum 5 and product 6.
For S=-3 and P=-10, x^2-Sx+P=x^2+3x-10. Brackets around S prevent the sign error.
Form a polynomial with sum 1 and product -6, then find its zeroes.
Complete x^2-(__)x+(__) for (S,P)=(7,10), (-2,-15), and (0,-9). Then factorise to check one case.
Compare p(x)=x^2-5x+6 and q(x)=4x^2-20x+24. Factorise both, list their zeroes and explain why multiplying by 4 changes no zero.
A quadratic graph has zeroes -2 and 4. Write the factors and simplest polynomial. Explain why the graph's vertical scale could correspond to a non-zero multiple without changing those zeroes.
Show S and P, substitute into x^2-Sx+P, factorise the result and state both zeroes. This makes each stage visible for marking.
The sum and product are -1 and -12. Form the monic polynomial, find its zeroes, write a different polynomial with the same zeroes, and explain why both answers work.