If yu can answer these three yu get 100 points gots to be smart smart

A.

4x² - 26 x + 42 = 0

We'll factor out and cancel the 2,

2x² - 13x + 21 = 0

We have to factor, so the two constants will multiply to 21.  There aren't that many choices, 1*21 and 7*3 and their negatives.  We definitely want the negatives because of the minus sign in -13x.  Trying a few things gives

(2x - 7)(x - 3) = 0

Solving,

x = 7/2 or x=3

In complete sentences, this means when we evaluate 4x² - 26 x + 42 at x=7/2 we'll get zero.  Same goes for x=3.

B.

(-2x³ + x - 5)(x³ -3x - 4)

By the distributive law,

= -2x³(x³ -3x - 4) + x(x³ -3x - 4) - 5(x³ -3x - 4)

We distribute again, three times,

= -2x⁶  + 6x⁴ + 8x³ + x⁴ - 3x² - 4x - 5x³ + 15x + 20

Then we collect like terms, for example 6x⁴ +  x⁴ = 7x⁴.

=  -2x⁶ + 7x⁴ + 3x³ - 3x² + 11x + 20

That's it.  In complete sentences we multiplied the trinomials by using the distributive law of multiplication over addition four times and then collecting like terms.

C.

(2x - 4)(3x² - x + 4)

= 2x(3x² - x + 4)  - 4(3x² - x + 4)

= 6x³ - 2x² + 8x - 12x² + 4x - 16

= 6x³- 14x² + 12x - 16

b. No, not equal, 2x-4 and 4x-2 are two different factors.

1) 2(x - 3)(2x - 7) = 0

The quadratic cuts the x-axis at x=3 and x=3.5

2) -2x⁶ + 7x⁴ + 3x³ - 3x² + 11x + 20

3) 6x³ - 14x² + 12x - 16; No

Step-by-step explanation:

4x² - 26x + 42 = 0

2(2x² - 13x + 21) = 0

2(2x² - 6x - 7x + 21) = 0

2[2x(x - 3) - 7(x - 3)] = 0

2(x - 3)(2x - 7) = 0

x = 3, 3.5

Means the quadratic cuts the x-axis at 3 and 3.5

(-2x³ + x - 5)(x³ - 3x - 4)

-2x⁶ + x⁴ - 5x³ + 6x⁴ - 3x² + 15x + 8x³ - 4x + 20

= -2x⁶ + 7x⁴ + 3x³ - 3x² + 11x + 20

To get this

I multiplied each term in the second bracket with all the terms in the first bracket. Then added all like terms.

(2x - 4)(3x² - x + 4)

6x³ - 2x² + 8x - 12x² + 4x - 16

= 6x³ - 14x² + 12x - 16

(4x - 2)(3x² - x + 4)

Will be a different expression

The term in x³ here would be:

4x×3x² = 12x³

Similarly all other terms are going to be different too

its  b

Step-by-step explanation:

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