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3 years ago

F(x)=15x³+22x²-15x+2 Write f(x) as a product of linear factors.

Mathematics

1 answer:

Goryan [66]3 years ago

3 0

Answer:

$(5x - 1)(3x - 1)(x + 2)$

Step-by-step explanation:

$(15 {x}^{3} + 22 {x}^{2} - 15x + 2)$

Apply Rational Root Theorem, our possible roots will be

plus or minus( 2/15, 2/5,2/3,2, 1/15,1/5,1/3,1).

I tried root -2 and it work so

If we apply synthetic dividon, we would be left with

$15 {x}^{2} - 8x + 1$

We can factor this regularly.

Apply AC method that a number

AC will multiply to 15 but add to -8.

The answer are -5 and -3 so we write this as

$15 {x}^{2} - 5x - 3x + 1$

Factor by grouping

$(15x {}^{2} - 5x) - (3x + 1)$

$5x(3x - 1) - 1(3x - 1)$

So our factor are

$(5x - 1)(3x - 1)(x + 2)$

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kherson [118]

Answer:

$25.57

Step-by-step explanation:

First, you have to multiply $5.29 by 2 because she wants to spend $5.29 on 2 pairs of socks.

5.29 * 2 = $10.58

Then you want to add $10.58 and $14.99 together.

$10.58 + $14.99 = $25.57

3 0

3 years ago

Calculate the speed of a car that went a distance of 125 miles in 2 hours<br> time.

DerKrebs [107]

Answer:

62.5

Step-by-step explanation:

speed = distance / time taken

= 125/2

=62.5miles per hour

5 0

3 years ago

To make cheesecake you need about 21/2pounds of creamcheese how many pounds of cream cheese will you need to make 2 cheesecakes

Elina [12.6K]

2 1/2 pounds of cream cheese for a cheesecake
to find the amount needed to make 2 cheesecakes multiply 2 1/2 times 2
2 1/2 times 2 = 5
you need 5 pounds of cream cheese to make 2 cheesecakes

6 0

3 years ago

An exterior angle of an isosceles triangle has measure 130°. Find two possible sets of measures for the angles of the triangle.

frozen [14]

(1) it can be both exterior of vertex or base. 180-130=50°
vertex is 50°: base=(180-50)/2=65°
base is 50°: vertex= 180-2*50=80°

vertex 50 base 65; vertex 80, base 50

(2)base=180-130=50°
vertex=180-2*50=80°

base 50 vertex 80

5 0

3 years ago

. For each of these intervals, list all its elements or explain why it is empty. a) [a, a] b) [a, a) c) (a, a] d) (a, a) e) (a,

Eva8 [605]

Answer:

Elements are of the form

(i) $[a,a]=\{[x,y] : a\leq x\leq a, a\leq y\leq a; a\in \mathbb R\}$

(ii) $[a,b)=\{[x,y) :a\leq x$

(iii) $(a,a]=\{(x,y] :a$

(iv) $(a,a)=\{(x,y): a$

(v) (a,b) where a>b= $\{(x,y) : a>x>b,a>y>b;a>b,a,b \in \mathbb R\}$

(vi) [a,b] where a>b= $\{[x,y] : a\geq x\geq b,a\geq y\geq b;a>b,a,b \in \mathbb R\}$

Step-by-step explanation:

Given intervals are,

(i) [a,a] (ii) [a,a) (iii) (a,a] (iv) (a,a) (v) (a,b) where a>b (vi) [a,b] where a>b.

To show all its elements,

(i) [a,a]

Imply the set including aa from left as well as right side.

Its elements are of the form.

$\{[a,a] : a\in \mathbb R\}=\{[0,0],[1, 1],[-1,-1],[2,2],[-2,-2],[3,3],[-3,-3],........\}$

Since there is a singleton element a of real numbers, this set is empty.

Because there is no increment so if $a\in \mathbb R$ then the set [a,a] represents singleton sets, and singleton sets are empty so is [a,a].

(ii) [a,a)

This means given interval containing a by left and exclude a by right.

Its elements are of the form.

$[ 1, 1),[-1,-1),[2,2),[-2,-2),[3,3),[-3,-3),........$

Since there is a singleton element a of real numbers withis the set, this set is empty.

Because there is no increment so if $a\in \mathbb R$ then the set [a,a) represents singleton sets, and singleton sets are empty so is [a.a).

(iii) (a,a]

It means the interval not taking a by left and include a by right.

Its elements are of the form.

$( 1, 1],(-1,-1],(2,2],(-2,-2],(3,3],(-3,-3],........$

Since there is a singleton element a of real numbers, this set is empty.

Because there is no increment so if $a\in \mathbb R$ then the set (a,a] represents singleton sets, and singleton sets are empty so is (a,a].

(iv) (a,a)

Means given set excluding a by left as well as right.

Since there is a singleton element a of real numbers, this set is empty.

Its elements are of the form.

$( 1, 1),(-1,-1],(2,2],(-2,-2],(3,3],(-3,-3],........$

Because there is no increment so if $a\in \mathbb R$ then the set (a,a) represents singleton sets, and singleton sets are empty, so is (a,a).

(v) (a,b) where a>b.

Which indicate the interval containing a, b such that increment of x is always greater than increment of y which not take x and y by any side of the interval.

That is the graph is bounded by value of a and it contains elements like it we fixed a=5 then,

$(a,b)=\{(5,0),(5,1),(5,2).....\}$ e.t.c

So this set is connected and we know singletons are connected in $\mathbb R$ . Hence given set is empty.

(vi) [a,b] where $a\leq b$ .

Which indicate the interval containing a, b such that increment of x is always greater than increment of y which include both x and y.

That is the graph is bounded by value of a and it contains elements like it we fixed a=5 then,

$[a,b]=\{[5,0],[5,1],[5,2].....\}$ e.t.c

So this set is connected and we know singletons are connected in $\mathbb R$ . Hence given set is empty.

8 0

3 years ago