All categories

3 years ago

Factor -7x^3+21x^2+3x-9 by grouping. what is the resulting expression?

Mathematics

1 answer:

Daniel [21]3 years ago

8 0

Answer:

c) (3-7x^2)(x-3)

Step-by-step explanation:

-7x^3+21x^2+3x-9

Factor out a -7x^2 from the first two terms and a 3 from the last two terms

-7x^2 (x -3) +3(x-3)

Now factor out (x-3)

(x-3) (-7x^2+3)

Rearranging)

(x-3) (3-7x^2)

You might be interested in

Select the correct numerator 7/9 - ?/27

damaskus [11]

Answer:

Step-by-step explanation:

27÷3 is 9 and 21÷3 is 7

7 0

2 years ago

Value 23.7 is 1/10 of

inna [77]

Is 19:17 because u have to value the number of the first like this 78:807y :997

6 0

2 years ago

Here is the rest you could help me in any way

Mila [183]

Answer:

6:$255.50

7:$21.25

8:$219.50

9:$143.75

Step-by-step explanation:

Can i have brainiest?!!!!

4 0

2 years ago

Read 2 more answers

Solve for x in IJK Algebra

Oksi-84 [34.3K]

Answer:

x = 60

Step-by-step explanation:

The sum of the angles in a triangle is 180

40 + x+20 + x = 180

Combine like terms

2x+60 =180

Subtract 60 from each side

2x+60-60 =180-60

2x =120

Divide by 2

2x/2 =120/2

x = 60

5 0

2 years ago

Read 2 more answers

Someone help me out please

uysha [10]

1 step (B): raise both sides of the equation to the power of 2.

$(\sqrt{x+3}-\sqrt{2x-1})^2=(-2)^2,\\ (x+3)-2\sqrt{x+3}\cdot \sqrt{2x-1}+(2x-1)=4,\\ 3x+2-2\sqrt{x+3}\cdot \sqrt{2x-1}=4$ .

2 step (A): simplify to obtain the final radical term on one side of the equation.

$-2\sqrt{x+3}\cdot \sqrt{2x-1}=4-3x-2,\\ -2\sqrt{x+3}\cdot \sqrt{2x-1}=2-3x,\\ 2\sqrt{x+3}\cdot \sqrt{2x-1}=3x-2$ .

3 step (F): raise both sides of the equation to the power of 2 again.

$(2\sqrt{x+3}\cdot \sqrt{2x-1})^2=(3x-2)^2,\\ 4(x+3)(2x-1)=(3x-2)^2$ .

4 step (E): simplify to get a quadratic equation.

$4(2x^2-x+6x-3)=(3x)^2-2\cdot 3x\cdot 2+2^2,\\ 8x^2+20x-12=9x^2-12x+4,\\ x^2-32x+16=0$ .

5 step (D): use the quadratic formula to find the values of x.

$D=(-32)^2-4\cdot 16=1024-64=960, \\ \sqrt{D} =8\sqrt{5} ,\\ x_{1,2}=\dfrac{32\pm 8\sqrt{5}}{2} =16\pm 4\sqrt{5}$ .

6 step (C): apply the zero product rule.

$x^2-32x+16=(x-16-4\sqrt{5}) (x-16+4\sqrt{5}) ,\\ (x-16-4\sqrt{5}) (x-16+4\sqrt{5}) =0,\\ x_1=16+4\sqrt{5} ,x_2=16-4\sqrt{5}$ .

Additional 7 step: check these solutions, substituting into the initial equation.

3 0

2 years ago