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1 year ago

Expand: f(x) = x4 − x3 + x2 + x −

Mathematics

2 answers:

lilavasa [31]1 year ago

8 0

Answer:

Step-by-step explanation:

its right i got it on edge 2020

nikitadnepr [17]1 year ago

6 0

Answer:

(x+3)(x+2)

Step-by-step explanation:

You might be interested in

What are the zeros of the quadratic function f(x) = 2x2 + 8x – 3?

jenyasd209 [6]

Answer:

$x=-2-\sqrt{\dfrac{11}{2}}\ \text{and}\ x=-2+\sqrt{\dfrac{11}{2}}$

Step-by-step explanation:

My favorite way to go at this is to look at a graph. It shows the vertex at (-2, -11). Since the leading coefficient is 2, this means the roots are ...

$-2\pm\sqrt{\dfrac{11}{2}}$

where the 2 in the denominator of the radical is the leading coefficient.

You can also use other clues:

the axis of symmetry is -b/(2a) = -8/(2(2)) = -2, so answer choices C and D don't work
the single change in sign in the coefficients (+ + -) tells you there is one positive real root, so answer choice B doesn't work.

The first answer choice is the only one with values symmetrical about -2 and one of them positive.

You may be expected to use the quadratic formula:

$x=\dfrac{-b\pm\sqrt{b^-4ac}}{2a}=\dfrac{-8\pm\sqrt{8^2-4(2)(-3)}}{2(2)}\\\\=\dfrac{-8}{4}\pm\dfrac{\sqrt{88}}{4}=-2\pm\sqrt{\dfrac{11}{2}}$

7 0

2 years ago

Help please!!!! real answers please need it in 1 hour.

s2008m [1.1K]

Answer:

5m + 37 = 79.50

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Micaela has a bat that is 30 inches long. She wants to fit it in a cube shaped box that has sides that are 17 inches. Will the b

valentina_108 [34]

No because 30 divided by 17 = 1.765

6 0

2 years ago

Read 2 more answers

Let P and Q be polynomials with positive coefficients. Consider the limit below. lim x→[infinity] P(x) Q(x) (a) Find the limit i

jenyasd209 [6]

Answer:

If the limit that you want to find is $\lim_{x\to \infty}\dfrac{P(x)}{Q(x)}$ then you can use the following proof.

Step-by-step explanation:

Let $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}$ and $Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0}$ be the given polinomials. Then

$\dfrac{P(x)}{Q(x)}=\dfrac{x^{n}(a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n})}{x^{m}(b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m})}=x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}$

Observe that

$\lim_{x\to \infty}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\dfrac{a_{n}}{b_{m}}$

and

$\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}$

Then

$\lim_{x\to \infty}=\lim_{x\to \infty}x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\begin{cases}0 & \text{if}\,\, nm \end{cases}$

3 0

2 years ago

Evaluate ∣∣256+y∣∣ for y=74. A. 225 B. 315 C. 345 D. 4712

aleksklad [387]

Answer:

678

Step-by-step explanation:

5 0

2 years ago