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4 years ago

All algebraic formula with proof its urgent do for brainlist

Mathematics

1 answer:

Gnom [1K]4 years ago

8 0

Here is a list of Algebraic formulas –

a2 – b2 = (a – b)(a + b)

(a+b)2 = a2 + 2ab + b2

a2 + b2 = (a – b)2 + 2ab

(a – b)2 = a2 – 2ab + b2

(a + b + c)2 = a2 + b2 + c2 + 2ab + 2ac + 2bc

(a – b – c)2 = a2 + b2 + c2 – 2ab – 2ac + 2bc

(a + b)3 = a3 + 3a2b + 3ab2 + b3 ; (a + b)3 = a3 + b3 + 3ab(a + b)

(a – b)3 = a3 – 3a2b + 3ab2 – b3

a3 – b3 = (a – b)(a2 + ab + b2)

a3 + b3 = (a + b)(a2 – ab + b2)

(a + b)3 = a3 + 3a2b + 3ab2 + b3

(a – b)3 = a3 – 3a2b + 3ab2 – b3

(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4)

(a – b)4 = a4 – 4a3b + 6a2b2 – 4ab3 + b4)

a4 – b4 = (a – b)(a + b)(a2 + b2)

a5 – b5 = (a – b)(a4 + a3b + a2b2 + ab3 + b4)

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Part 2:

castortr0y [4]

Answer:

2. ±1, ±2, ±3, ±4, ±6, ±9, ±12, ±18, ±36

3. see below for synthetic division

4. complex roots: ±3i

5. see the second attachment for a graph (has no symmetry)

Step-by-step explanation:

2. The Rational Root Theorem tells you possible rational roots are the form of ...

±(divisor of constant term) / (divisor of leading coefficient)

Here, we're lucky in that the leading coefficient is 1. The constant is 36, and its divisors are 1, 2, 3, 4, 6, 9, 12, 18, 36. Possible rational roots, according to the Rational Root Theorem are ...

±1, ±2, ±3, ±4, ±6, ±9, ±12, ±18, ±36

3. When actually searching for possible roots, it is useful to narrow down this list of 18 candidates. We can do this using Descarte's Rule of Signs and a couple of other simple tests. The rule of signs has us look at the signs of the coefficients. They are ...

+ + + + +

There are no sign changes, hence no positive real roots. This means we're looking at negative roots only.

We know that p(0) = 36, the constant term. We can evaluate p(-1) simply, by reversing the signs of the odd-degree coefficients and adding them up:

1 -4 +13 -36 +36 = 10

So p(0) = 36 and p(-1) = 10. The average rate of change is -26, so we expect p(-2) to be zero or less. That value (-2) may be a good first choice for a possible root.

The synthetic division for the root -2 is shown in the first attachment. It shows a remainder of 0, so -2 is a root. The polynomial with the corresponding factor divided out is ...

q(x) = x^3 +2x^2 +9x +18

We can apply the above steps to this polynomial. It will have no positive real roots. The value of q(0) is 18, and the value of q(-1) is -1 +2 -9 +18 = 10. The average rate of change between these points is (18 -10)/(0 -(-1)) = 8, so we expect the function to be zero or negative between x=-2 and x=-3.

Further, we observe that the ratio of the first two coefficients is 1 : 2, the same as the 9 : 18 ratio of the last two coefficients. This suggests we can factor this polynomial by "grouping."

q(x) = (x^3 +2x^2) +(9x +18) = x^2(x +2) +9(x +2) = (x^2 +9)(x +2)

So, without doing any more synthetic division, we have found a second root at x = -2. This means -2 is a double root.

Our factorization so far is ...

p(x) = (x +2)^2(x^2 +9)

4. The remaining polynomial factor (x^2 +9) has only complex roots:

x^2 +9 = 0

x^2 = -9

x = ±√(-9)

x = ±3i . . . . the complex roots

5. The graph is shown in the second attachment, along with the graph of the final quadratic factor.

The leading coefficient is positive and p(x) is of even degree, so it will have a general U shape, going to +∞ for large positive and negative values of x.

Since x = -2 is a double root, the graph touches the x-axis there, but does not cross it.

The graph of p(x) shows no symmetry.

______

Comment on the graph

With modern graphing calculator software it is entirely feasible to start by graphing the polynomial, then reading the (rational) roots from the graph. The same calculator can effectively divide out those roots. If the result is a quadratic, its vertex form equation can be read from the graph, making it easy to find any additional real or complex roots.

This graph shows ...

p(x) = (x +2)^2·q(x)

q(x) = x^2 +9