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3 years ago

6

PLEASE HELP ME!!!!! Factor completely b^4 − b^3 + b − 1

1 answer:

Afina-wow [57]3 years ago

7 0

<span>b^4 − b^3 + b − 1
=(</span><span>b^4 − b^3) + (b − 1)
= b^3(b - 1) + (b - 1)
=(b - 1)(b^3 + 1)</span>

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Is the relation depicted in the chart below a function?

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1 year ago

Read 2 more answers

Given the equation f(x)=2x^2-14x+10, find g(x), the image of f(x) after a ry=x (reflection over the line y=x). Express your answ

Illusion [34]

Given:

The function is:

$f(x)=2x^2-14x+10$

The function g(x), the image of f(x) after a $r_{y=x}$ (reflection over the line y=x).

To find:

The function g(x).

Solution:

We have,

$f(x)=2x^2-14x+10$

Substitute $f(x)=y$ in the given function.

$y=2x^2-14x+10$

The function g(x), the image of f(x) after a $r_{y=x}$ (reflection over the line y=x). So, interchange x and y.

$x=2y^2-14y+10$

Now, we need to find the value of y.

$x=2(y^2-7y+5)$

$\dfrac{x}{2}=y^2-7y+5$ [Divide both sides by 2]

$\dfrac{x}{2}-5=y^2-7y$ [Subtract 5 from both sides]

Add both sides half of square of coefficient of y, i.e. $(\dfrac{-7}{2})^2$ , to make it perfect square.

$\dfrac{x}{2}-5+(\dfrac{-7}{2})^2=y^2-7y+(\dfrac{-7}{2})^2$

$\dfrac{x}{2}-5+\dfrac{49}{4}=y^2-7y+(\dfrac{7}{2})^2$

$\dfrac{x}{2}-5+\dfrac{49}{4}=\left(y-\dfrac{7}{2}\right)^2$ $[\because (a-b)^2=a^2-2ab+b^2]$

$\dfrac{x}{2}+\dfrac{49-20}{4}=\left(y-\dfrac{7}{2}\right)^2$

Taking square root on both sides, we get

$\pm\sqrt{\dfrac{x}{2}+\dfrac{29}{4}}=y-\dfrac{7}{2}$

$\dfrac{7}{2}\pm\sqrt{\dfrac{2x+29}{4}}=y$

$\dfrac{7}{2}\pm\dfrac{\sqrt{2x+29}}{2}=y$

$\dfrac{7\pm \sqrt{2x+29}}{2}=y$

Substituting $y=g(x)$ , we get

$\dfrac{7\pm \sqrt{2x+29}}{2}=g(x)$

Therefore, the required function is $g(x)=\dfrac{7\pm \sqrt{2x+29}}{2}$ .

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2 years ago