factoring (algebra 2)?

Question

Suppose a quadratic function f(x) = x^2 + bx + c has zeros p and q.

a) Let k be an integer. Using only b, c, and k (and x), write a new qudratic function whose zeros are p+k and q+k.

b) Repeat part (a), but write a new function whose zeros are p-k and q-k.






Let k represent a positive integer, so that 2k represents a positive even integer, and suppose we want to answer the following question: How can 2k be given as a sum of two numbers x a nd y so that xy is as large as possible?

a) Answer the question specifically for 2k=10 by trying all possible pairs of integers whose sum is 10. Based on your answer, make a conjecture for the general case.

b. Prove the conjecture you made in part (a) by writing a quadratic function in factored form.

Answer 1

1a.) Remember that a quadratic equation can be expressed as x^2 + bx + c or (x + p)(x + q).

Therefore:
c = pq
b = p+q

So, (x + (p + k)) * (x + (q + k)) = x^2 + (p + k)x + (q + k)x + (p + k) (q + k)
Distribute:
= x^2 + (p + q + 2k)x + pq + pk + qk + k^2
Combine like terms:
= x^2 + (p + q + 2k)x + c + (p + q)k + k^2
Substitute c and b:
= x^2 + (b + 2k)x + c + bk + k^2 (solution!)

b.) So, (x + (p - k)) * (x + (q - k)) = x^2 + (p - k)x + (q - k)x + (p - k) (q - k)
Distribute:
= x^2 + (p + q - 2k)x + pq - pk - qk + k^2
Combine like terms:
= x^2 + (p + q - 2k)x + c - (p + q)k + k^2
Substitute c and b:
= x^2 + (b - 2k)x + c - bk + k^2 (solution!)

2.) 2k = 10. 10 may equal: 1 + 9 (1*9 = 9), 2 + 8 (16), 3 + 7 (21), 4 + 6 (24), 5 + 5 (25). The conjecture is that xy will be greatest when x = y.

From that, you can do b.

Answer 2

i don't know if this website might help, but you can try it.
http://algebrahelp.com/calculators/equation/