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Phantasy [73]
3 years ago
5

Factor 4x^2 + 12x + 5

4x - 5" alt="12x^{2} - 4x - 5" align="absmiddle" class="latex-formula">
​
Mathematics
1 answer:
zheka24 [161]3 years ago
3 0

Answer:

(2x + 5)(2x + 1).

(6x - 5)(2x + 1).

Step-by-step explanation:

4x^2 + 12x + 5.

Use the 'ac' method:

a * c = 4 * 5 = 20  so we need 2 numbers whose product is 20 and whose sum is +12.:-  2 and 10 look good, so we write:

4x^2 + 2x + 10x + 5

Now we factor by grouping:-

= 2x(2x + 1) + 5(2x + 1)

= (2x + 5)(2x + 1)   (answer).

12x^2  - 4x  - 5

ac = 12*-5 = - 60  and  we need 2 numbers sum to -4 .  these numbers could be  -10 and + 6:

12x^2 + 6x - 10x - 5

= 6x(2x + 1) - 5(2x + 1)

= (6x - 5)(2x + 1) (answer).

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Step-by-step explanation:

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~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ R(\stackrel{x_1}{-7}~,~\stackrel{y_1}{5})\qquad S(\stackrel{x_2}{17}~,~\stackrel{y_2}{5}) ~\hfill RS=\sqrt{(~~ 17- (-7)~~)^2 + (~~ 5- 5~~)^2} \\\\\\ ~\hfill RS=\sqrt{( 24)^2 + ( 0)^2}\implies \boxed{RS=24} \\\\\\ S(\stackrel{x_1}{17}~,~\stackrel{y_1}{5})\qquad T(\stackrel{x_2}{5}~,~\stackrel{y_2}{0}) ~\hfill ST=\sqrt{(~~ 5- 17~~)^2 + (~~ 0- 5~~)^2}

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