Download Exam correctives ch 6

Exam Correctives
Honors Algebra II
Chapter 6
problem 1: Write in standard form.
Then classify the polynomial by degree and
by number of terms.
a) x(x ! 1)(5x + 4)
b) x 2 (x ! 7)(2x + 3)
c) x(2x ! 3)2
a) 4x + 8x ! 96x
2
b) 3x ! 6x ! 45x
3
2
c) 6x ! 54x + 108x
3
a) x 3 + 216
b) 2x 3 ! 54
c) 4x 3 ! 500
problem 8: Find the roots of the
problem 2: Write in factored form.
3
problem 7: Factor the expression.
2
problem 3: Write a polynomial
equation in standard form with zeros at:
a) 2,1,5
b) 1, !3, 4
c) 6, !2, 2
problem 4: Find the zeros and state
their multiplicity.
a) f (x) = x(x ! 9)4 (x + 2)5
b) f (x) = x 2 (x + 5)3 (x ! 4)
c) f (x) = x 3 (x ! 6)5 (x ! 8)2
problem 5: Divide:
a) 4x 3 + 3x 2 ! x ! 3 by x + 2
b) 3x 3 ! 2x 2 ! x ! 4 by x ! 2
c) -x 3 + 2x 2 ! 4x ! 4 by x ! 4
problem 6: Use synthetic division and
the Remainder Theorem to find:
a) p(!3) p(x) = x 4 ! 4x 3 + 4x 2 ! 9x ! 2
equation.
a) 125x 3 + 343 = 0
b) x 3 + 27 = 0
c) 27x 3 + 125 = 0
problem 9: Solve.
a) x 4 ! 34x 2 = !225
b) x 4 ! 4x 2 = 60
c) x 4 + 21x 2 = 100
problem 10: Find all the zeros of the
function.
a) x 3 + x 2 ! 2x ! 2 = 0
b) x 3 ! x 2 + 16x ! 16 = 0
c) x 3 ! 3x 2 + x ! 3 = 0
problem 11: Find a third-degree
polynomial equation with rational
coefficients that has roots:
a) ! 2 & 5 + i
b) ! 4 & 2 + i
c) ! 6 & 2 + i
problem 12: Find the number of
complex roots and the possible number of
real roots for:
b) p(4) p(x) = x 4 ! x 3 ! 7x 2 + 9x ! 3
a) 9x 8 + x 3 ! 2x 2 ! 2 = 0
c) p(!4) p(x) = x 4 + 7x 3 + x 2 ! 3x ! 6
b) 16x 11 ! x 8 + 16 = 0
c) ! 4x 12 ! 3x 6 + x ! 3 = 0