term 4
= C(6,3)(2x)^3 (3y)^3
= 20(8x^3)(27y^3
= 4320x^3 y^3
2. hard to show synthetic division here and make the numbers line up properly,
in effect you are dividing by x+1
I got -8
3. I noticed that 4096 is a perfect square
x^6 + 128x^3 + 4096 = 0
(x^3 + 64)^2 = 0
x^3 + 64 = 0
x^3 = -64
x = -4
4.
(t-2)(t+1)(t-1)
= (t+2)(t^2 - 1)
= t^3 - t + 2t^2 - 2
= t^3 + 2t^2 - t - 2
what do you think?
1. Find the 4th term in the binomial expansion.(2x + 3y)^6
2. Use synthetic division and the remainder theorem to find P(a). P(x)= x^5 + 3x^4 + 3x − 7; P(-1)
3. Factor the expression on the left side of the equation. Then solve the equation. x^6 + 128x^3 + 4096 = 0
4.) Write the polynomial in standard form. Then classify it by degree and number of terms.(t - 2)(t +1)(t - 1)
2 answers
#1
since (a+b)^6 = a^6 + 6a^5b + ...
if we let a=2x and b=3y we get
(2x)^6 + 6(2x)^5(3y) + 15(2x)^4(3y)^2 + 20(2x)^3(3y)^3 + 15(2x)^2(3y)^4 + 6(2x)(3y)^5 + (3y)^6
64x^6 + 576x^5y + 2160x^4y^2 + 4320x^3y^3 + 4860x^2y^4 + 2916xy^5 + 729y^6
#2
Hard to show. The bottom-line entries are 1 2 -2 2 1 -8
There is a nice calculator at
http://www.mathportal.org/calculators/polynomials-solvers/synthetic-division-calculator.php
#3
since 4096=64^2 and 128=2*64, we have
(x^3+64)^2
#4
Just looking he factors you know it is degree 3
Its expanded form is
x^3 - 2t^2 - t + 2
I expect you can count the terms...
since (a+b)^6 = a^6 + 6a^5b + ...
if we let a=2x and b=3y we get
(2x)^6 + 6(2x)^5(3y) + 15(2x)^4(3y)^2 + 20(2x)^3(3y)^3 + 15(2x)^2(3y)^4 + 6(2x)(3y)^5 + (3y)^6
64x^6 + 576x^5y + 2160x^4y^2 + 4320x^3y^3 + 4860x^2y^4 + 2916xy^5 + 729y^6
#2
Hard to show. The bottom-line entries are 1 2 -2 2 1 -8
There is a nice calculator at
http://www.mathportal.org/calculators/polynomials-solvers/synthetic-division-calculator.php
#3
since 4096=64^2 and 128=2*64, we have
(x^3+64)^2
#4
Just looking he factors you know it is degree 3
Its expanded form is
x^3 - 2t^2 - t + 2
I expect you can count the terms...
1 answer