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4 years ago

match each binomial expression with the set of coefficient of the term obtain by expanding the expression

Mathematics

2 answers:

mars1129 [50]4 years ago

8 0

$(2x+y)^3$ = $8x^3+12x^2y+6xy^2+y^3$

The coefficients are 8 , 12, 6, 1

$(2x+3y)^4$ = $16x^4+96x^3y+216x^2y^2+216xy^3+81y^4$

The coefficients are 16, 96, 216, 216, 81

$(3x+2y)^3$ = $27x^3+54x^2y+36xy^2+8y^3$

The coefficients are 27, 54,36 , 8

$(x+y)^4$ = x^4+4x^3y+6x^2y^2+4xy^3+y^4

The coefficients are 1, 4, 6, 4, 1

anygoal [31]4 years ago

4 0

By expanding the term $(2x+y)^3$ , we get

$8x^3 + 12x^2y+6xy^2+y^3$ .

Therefore, the coefficients for $(2x+y)^3$ are 8, 12, 6, 1.

By expanding the term $(2x+3y)^4$ , we get

$16x^4+96x^3y+216x^2y^2+216xy^3+81y^4$ .

Therefore, the coefficients for $(2x+3y)^4$ are 16, 96, 216, 216, 81.

By expanding the term $(3x+2y)^3$ , we get

$27x^3+54x^2y+36xy^2+8y^3$ .

Therefore, the coefficients for $(3x+2y)^3$ are 27, 54, 36, 8.

By expanding the term $(x+y)^4$ , we get

$x^4+4x^3y+6x^2y^2+4xy^3+y^4$ .

Therefore, the coefficients for $(x+y)^4$ are 1, 4, 6, 4, 1.

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3 years ago

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Darya [45]

A. Write a function f(x) to represent the price after the 80% markup.
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3 years ago

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8_murik_8 [283]

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Step-by-step explanation:

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4 years ago

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patriot [66]

Answer:

$f(x) = (x-4)(x-\frac{1}{2})^2 = x^3-5x^2+4.25x -1$

Step-by-step explanation:

Recall, if we have a polynomial of the form $(x-a)^k \cdot (x-b)^m$ , then we say that a is a zero of multiplicity k and b is a zero of multiplicty m. For example, in the polynomial of the form $(x+5)^10(x-2)^3$ -5 is a zero of multiplicity 10 and 2 is a zero of multiplicity 3. If we want to know the degree of the polynomial, just add the multiplicity of both zeros (13 in our example).

In this case, we know that the degree of our polynomial should be at least 3(multiplicity 2 and multiplicity 1). So, lets take the polynomial of the form

$f(x)=(x-a)(x-b)^2)$ .

In here, a is a zero with multiplicity 1 and b is a zero with multiplicity 2. We are also given that

$f(0) = -1 = (0-a)(0-b)^2 = -a\codt b^2$ .

Which implies that $1 = a\cdot b^2$ . Since the square of any number is a positive number, it must happen that a>0. So, we have that

$b= \pm \sqrt[]{\frac{1}{a}}$ .

We can choose any value of a and solve for b. Let us choose a=4. So we can have b=1/2 or b=-1/2. Let's use b=1/2. So our polynomial would be

$f(x) = (x-4)(x-\frac{1}{2})^2 = x^3-5x^2+4.25x -1$

which we can easily check that f(0)=-1.

6 0

4 years ago