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Ad libitum [116K]
3 years ago
11

Express the polynomial below by nesting. f(x)=4x³+3x²+2x-4​

Mathematics
1 answer:
BabaBlast [244]3 years ago
4 0

Answer:

f(x)=4x^{3} +3x^{2} +2x-4 can be expressed in nested form as ((4x+3)x+2)x-4

Step-by-step explanation:

Considering the polynomial

f(x)=4x^{3} +3x^{2} +2x-4

Expressing the polynomial by nesting

So,

f(x)=4x^{3} +3x^{2} +2x-4

Step 1) Writing down the coefficient and one factor of x from the first term and add on the coefficient of the next term.

4x+3

Step 2) Enclose this term in brackets, multiply by x and add on the next coefficient.

(4x+3)x+2

Step 3) Repeating the previous step until all terms are within the equation.

So,

((4x+3)x+2)x-4

Therefore, f(x)=4x^{3} +3x^{2} +2x-4 can be expressed in nested form as ((4x+3)x+2)x-4

<em>Keywords:</em><em> polynomial, nesting form</em>

<em>Learn more polynomials from brainly.com/question/12240569</em>

<em>#learnwithBrainly</em>

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Answer:

Part a) The new rectangle labeled in the attached figure N 2

Part b) The diagram of the new rectangle with their areas  in the attached figure N 3, and the trinomial is x^{2} +11x+28

Part c) The area of the second rectangle is 54 in^2

Part d) see the explanation

Step-by-step explanation:

The complete question in the attached figure N 1

Part a) If the original square is shown below with side lengths marked with x, label the second diagram to represent the new rectangle constructed by increasing the sides as described above

we know that

The dimensions of the new rectangle will be

Length=(x+4)\ in

width=(x+7)\ in

The diagram of the new rectangle in the attached figure N 2

Part b) Label each portion of the second diagram with their areas in terms of x (when applicable) State the product of (x+4) and (x+7) as a trinomial

The diagram of the new rectangle with their areas  in the attached figure N 3

we have that

To find out the area of each portion, multiply its length by its width

A1=(x)(x)=x^{2}\ in^2

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The total area of the second rectangle is the sum of the four areas

A=A1+A2+A3+A4

State the product of (x+4) and (x+7) as a trinomial

(x+4)(x+7)=x^{2}+7x+4x+28=x^{2} +11x+28

Part c) If the original square had a side length of  x = 2 inches, then what is the area of the  second rectangle?

we know that

The area of the second rectangle is equal to

A=A1+A2+A3+A4

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A2=(4)(2)=8\ in^2

A3=(2)(7)=14\ in^2

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A=4+8+14+28

A=54\ in^2

Part d) Verify that the trinomial you found in Part b) has the same value as Part c) for x=2 in

We have that

The trinomial is

A(x)=x^{2} +11x+28

For x=2 in

substitute and solve for A(x)

A(2)=2^{2} +11(2)+28

A(2)=4 +22+28

A(2)=54\ in^2 ----> verified

therefore

The trinomial represent the total area of the second rectangle

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