Write the quadratic equation whose roots are -4 and 3 and whose leading coefficient is 3

Answer 1

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When you have two roots and a desired first coefficient, you need to construct it by multiplying

(first coefficient)(x-root1)(x-root2) = 0

So substitute the roots and initial coefficient

3(x-3)(x-(-4)) = 0

3(x-3)(x+4) = 0

multiply it out

3 * (x²+x-12) = 3x² + 3x - 36 = 0

Substitute one of the roots (3) for x and verify that 0=0, and do the same by substituting -4 for x.

Answer 2

Polynomial: ax^2 + bx + c

Product of roots of polynomial = c/a

Sum of roots of polynomial = -b/a

a= 3

product of roots = -4 * 3 = -12

c/a = c/3 = -12 ---> c = 36

-b/a = -b/3 = -4 + 3 = -1 ---> b = 3

3x^2 + 3x - 12