Ask question

Ask question

All categories

2 years ago

5

Your friend attempted to factor an expression as shown. Find the error in your friends work. Then factor the expression correctl

y. Show work please.
2x^2-7x+5

2x^2-5x-2x+5

x(2x-5)+(2x-5)

(x+1)(2x-5)

1 answer:

loris [4]2 years ago

3 0

Answer:

Third step is incorrect. The correct factored form is (x-1)(2x-5).

Step-by-step explanation:

The given expression is

$2x^2-7x+5$

We need to find the factored form of this expression.

Step 1: Given

$2x^2-7x+5$

Step 2: Splitting the middle term method, the middle term can be written as (-5x-2x).

$2x^2-5x-2x+5$

$(2x^2-5x)+(-2x+5)$

Step 3: Taking out common factors from each parenthesis.

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

Step 4: Taking out common factors.

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

Therefore, the third step is incorrect. The correct factored form is (x-1)(2x-5).

You might be interested in

Which one is right..

frez [133]

Answer:

the answer would be C

Step-by-step explanation:

8 0

3 years ago

Prove that $5^{3^n} + 1$ is divisible by $3^{n + 1}$ for all nonnegative integers $n.$

Viktor [21]

When $n=0$ , we have

$5^{3^0} + 1 = 5^1 + 1 = 6$

$3^{0 + 1} = 3^1 = 3$

and of course 3 | 6. ("3 divides 6", in case the notation is unfamiliar.)

Suppose this is true for $n=k$ , that

$3^{k + 1} \mid 5^{3^k} + 1$

Now for $n=k+1$ , we have

$5^{3^{k+1}} + 1 = 5^{3^k \times 3} + 1 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k}\right)^3 + 1^3 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k} + 1\right) \left(\left(5^{3^k}\right)^2 - 5^{3^k} + 1\right)$

so we know the left side is at least divisible by $3^{k+1}$ by our assumption.

It remains to show that

$3 \mid \left(5^{3^k}\right)^2 - 5^{3^k} + 1$

which is easily done with Fermat's little theorem. It says

$a^p \equiv a \pmod p$

where $p$ is prime and $a$ is any integer. Then for any positive integer $x$ ,

$5^3 \equiv 5 \pmod 3 \implies (5^3)^x \equiv 5^x \pmod 3$

Furthermore,

$5^{3^k} \equiv 5^{3\times3^{k-1}} \equiv \left(5^{3^{k-1}}\right)^3 \equiv 5^{3^{k-1}} \pmod 3$

which goes all the way down to

$5^{3^k} \equiv 5 \pmod 3$

So, we find that

$\left(5^{3^k}\right)^2 - 5^{3^k} + 1 \equiv 5^2 - 5 + 1 \equiv 21 \equiv 0 \pmod3$

QED

5 0

1 year ago

Factor the polynomial

Anettt [7]

Answer:

(3x + 5)(9x^2 - 15x + 25)

Step-by-step explanation:

27x^3 + 125

= (3x)^3 + 5^3

= (3x + 5) ((3x)^2 - (3x)(5) + 5^2)

= (3x + 5)(9x^2 - 15x + 25)

FYI: a³ + b³ = (a + b)(a² – ab + b²)

8 0

1 year ago

Read 2 more answers

Hawlp giving brainlesst

Sedaia [141]

Answer:

B. $-\frac{7}{10}$ = -.7

Step-by-step explanation:

Replace all occurrences of − − with a single +. Two consecutive minus signs have the same mathematical meaning

$-.75+\frac{2}{5}+.4-\frac{3}{4}$

Find the common denominator.

$\frac{-.75*20}{20} +\frac{2*4}{20} +\frac{.4*20}{20}-\frac{3*5}{20}$

Simplify terms.

$\frac{-15+8+8-15}{20}$

Simplify the numerator.

$\frac{-14}{20}$

Reduce the expression by cancelling the common factors.

$-\frac{7}{10}$

<em>hope this helps, good luck :)</em>

4 0

3 years ago

Read 2 more answers

Todd has some folders he gave 2 to Tracy now he has 8 . how many folders did Todd have to begin with?

asambeis [7]

Answer: 10

Step-by-step explanation: 2+8=10

3 0

3 years ago

Read 2 more answers