How do you factor 7x^{2}+29x+4

1 answer:

7 0

When you have a coefficient in front of the "x^2" value, you have to multiply it by the end value (4)

so you want a set of integers that will add up to 29, by multiple to (7 x 4) = 28

The only numbers like that are 28 and 1, they add up to 29 and multiply to 29

Therefore, rewrite it like this 7x^2 + 28x + x + 4

and then factor out the 7x in the first two terms, to get this

7x(x+4) + x + 4, then factor out 1 in the latter two terms to get this:

7x(x+4) +1(x+4), then use grouping to combine what you've got:

The factors are: {{ (7x+1)(x+4) }}

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Step-by-step explanation:

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Find the perimeter of a field that has length 2/x + 1 and width 5/x^2 -1.

Semmy [17]

Answer:

D)4x + 6/(x + 1)(x - 1)

Step-by-step explanation:

A field is basically a rectangle, so to find the perimeter of our field we are using the formula for the perimeter of a rectangle

$p=2(l+w)$

where

$p$ is the perimeter

$l$ is the length

$w$ is the width

We know from our problem that the field has length 2/x + 1 and width 5/x^2 -1, so $l=\frac{2}{x+1}$ and $w=\frac{5}{x^2-1}$ .

Replacing values:

$p=2(l+w)$

$p=2(\frac{2}{x+1} +\frac{5}{x^2-1})$

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