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)
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
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. To find the 4th term in the binomial expansion of (2x + 3y)^6, we can use the binomial theorem formula. The formula is:
T(r+1) = nCr * (a^r) * (b^(n-r))
In this case, n is 6, r is 3 (since we want the 4th term), a is 2x, and b is 3y. Using this formula:
T(4) = 6C3 * (2x)^3 * (3y)^(6-3)
= 20 * 8x^3 * 27y^3
= 4320x^3y^3
Therefore, the 4th term in the binomial expansion is 4320x^3y^3.
2. To find P(-1) using synthetic division, we need to divide P(x) = x^5 + 3x^4 + 3x − 7 by (x - (-1)) or (x + 1).
Using synthetic division, we set up the division as follows:
-1 | 1 3 0 3 0 -7
Now, let's perform the synthetic division:
1 -1 1 -2 2 -2
________________________
1 2 -1 1 -2 4
The remainder is 4, so P(-1) = 4.
3. To factor the expression x^6 + 128x^3 + 4096 = 0, we can notice that it is a quadratic in terms of x^3. Therefore, let's substitute a variable, let's say u = x^3. This makes the equation become:
u^2 + 128u + 4096 = 0.
Now, we can factor this quadratic equation:
(u + 64)(u + 64) = 0.
Thus, we have (x^3 + 64)(x^3 + 64) = 0.
To solve this equation, we set each factor equal to zero:
x^3 + 64 = 0
Taking the cube root of both sides:
x = -4
Therefore, the solution to the equation x^6 + 128x^3 + 4096 = 0 is x = -4.
4. To write the polynomial (t - 2)(t + 1)(t - 1) in standard form, we need to multiply it out:
(t - 2)(t + 1)(t - 1) = (t^2 - t - 2)(t - 1)
= t^3 - t^2 - 2t - t^2 + t + 2
= t^3 - 2t^2 - t + 2.
The polynomial is of degree 3 (highest power of t is 3) and has 4 terms (since there are no like terms that can be combined, each parenthesis is multiplied out separately).
1. To find the 4th term in the binomial expansion of (2x + 3y)^6, we can use the binomial theorem. The binomial theorem states that the nth term in the expansion of (a + b)^n is given by the formula:
T_n = (n! / ((n-r)! * r!)) * (a^(n-r)) * (b^r)
In this case, a = 2x, b = 3y, and n = 6. We want to find the 4th term, so r = 4.
Plugging these values into the formula, we have:
T_4 = (6! / ((6-4)! * 4!)) * (2x)^(6-4) * (3y)^4
Simplifying further:
T_4 = (6! / (2! * 4!)) * (2x)^2 * (3y)^4
= (6 * 5 / 2 * 1) * (2x)^2 * (3y)^4
= 15 * 4x^2 * 81y^4
= 60x^2 * 81y^4
= 4860x^2y^4
Therefore, the 4th term in the binomial expansion of (2x + 3y)^6 is 4860x^2y^4.
2. To use synthetic division and the remainder theorem to find P(-1) for the polynomial P(x) = x^5 + 3x^4 + 3x − 7, we follow the steps below:
Step 1: Set up the synthetic division table by writing down the coefficients of the polynomial in descending order:
-1 │ 1 3 3 0 -7
Step 2: Bring down the coefficient of the highest power of x (1 in this case) into the last row of the synthetic division table.
-1 │ 1 3 3 0 -7
────────────
1
Step 3: Multiply the divisor (-1) by the previous result (1) and write the result above the next coefficient.
-1 │ 1 3 3 0 -7
────────────
1
──
-1
Step 4: Add the new result (-1) to the next coefficient (3) and write the sum below the dividing line.
-1 │ 1 3 3 0 -7
────────────
1
──
-1
-2
Step 5: Repeat steps 3 and 4 until you reach the constant term (-7).
-1 │ 1 3 3 0 -7
────────────
1
──
-1
-2
──
-5
-1
────
1
Step 6: The last number in the final row of the synthetic division table (1 in this case) is the remainder.
Therefore, when we evaluate P(-1), we get a remainder of 1.
3. To factor the expression x^6 + 128x^3 + 4096 = 0, we can begin by recognizing the equation as a quadratic in terms of x^3. Let's substitute x^3 with a variable, such as u:
u^2 + 128u + 4096 = 0
Now we can factor this quadratic equation. We can search for two numbers that multiply to give 4096 and add up to 128. In this case, 64 and 64 satisfy these conditions.
(u + 64)(u + 64) = 0
Now let's replace u with x^3:
(x^3 + 64)(x^3 + 64) = 0
We have factored the expression on the left side of the equation. Now we can solve for x by setting each factor equal to zero:
x^3 + 64 = 0
x^3 = -64
To find the cube root of -64, we know that -4 cubed is -64. Therefore, we have:
x = -4
So the solution to the equation x^6 + 128x^3 + 4096 = 0 is x = -4.
4. To write the polynomial (t - 2)(t + 1)(t - 1) in standard form and classify it by degree and number of terms:
First, we'll multiply the factors together:
(t - 2)(t + 1)(t - 1) = (t^2 - t - 2)(t - 1)
= t^3 - t^2 - 2t - t^2 + t + 2
= t^3 - 2t^2 - t + 2
Now we can rearrange the terms in descending order of degree:
t^3 - 2t^2 - t + 2
This is the polynomial in standard form.
By examining the polynomial, we can see that it is a cubic polynomial because the highest exponent is 3. Additionally, it has four terms. Therefore, we can classify it as a cubic polynomial with four terms.