Question

Алгебра 1)28-7у² 2) -11х²+22х-11 3) (у²-1)²-9 4) с²+6с-40

Answer 1

Ответ:

28-7у²=7(4-у²)=7(2-у)(2+у)

-11х²+22х - 11= - 1(х²-2х+1)= - 11(х-1)(х-1= -11(х-1)²

х³у+8у=у(х³+8)=у(х+2)(х²-2х+4)

(у²-1)²-9=(у²-1)² - 3²=(у²-1-3)(у²-1+3)=(у²-4)(у²+2)=(у-2)(у+2)(у²+2)

Answer 2

To simplify the given algebraic expressions, I'll provide the step-by-step solutions for each one:

1. 28 - 7у²

This expression does not factor further, but you can rewrite it as follows: 28 - 7у² = 7(4 - у²)

2. -11х² + 22х - 11

To factor this quadratic expression, you can first find the common factor, which is -11: -11х² + 22х - 11 = -11(х² - 2х + 1)

Now, we have a perfect square trinomial within the parentheses. Factoring it: -11(х² - 2х + 1) = -11(х - 1)²

3. (у² - 1)² - 9

First, expand the squared term inside the parentheses: (у² - 1)² = у⁴ - 2у² + 1

Now, substitute this into the original expression: (у² - 1)² - 9 = у⁴ - 2у² + 1 - 9

Combine like terms: у⁴ - 2у² - 8

The expression is now fully simplified.

4. с² + 6с - 40

To factor this quadratic expression, look for two numbers that multiply to -40 and add up to the coefficient of the middle term, which is 6.

The two numbers are 10 and -4, as 10 * -4 = -40, and 10 + (-4) = 6.

Now, rewrite the middle term using these numbers and then factor by grouping: с² + 10с - 4с - 40

Factor by grouping: с(c + 10) - 4(c + 10)

Now, you can factor out the common binomial factor (c + 10): (c + 10)(с - 4)

So, the fully factored expression is (c + 10)(с - 4).

That's it! The given algebraic expressions are now simplified and factored where possible.