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AleksandrR [38]
4 years ago
11

How can I solve y^2 + 13y + 12?

Mathematics
2 answers:
zavuch27 [327]4 years ago
8 0
(y+1)(y+12) that’s what I got.
lidiya [134]4 years ago
7 0

(y + 1)(y + 12)

\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\orange{:}}}}}

{y}^{2}  + 13y + 12 \\  =  {y}^{2}  + 12y + 1y + 12 \\  taking \: common \: factor \: from \: the \: two \: pairs \\ = y(y + 12) + 1(y + 12) \\  = (y + 1)(y + 12)

\large\mathfrak{{\pmb{\underline{\orange{Happy\:learning }}{\orange{.}}}}}

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What is the quotient (x^3+8)/(x+2)
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Answer:

Step-by-step explanation:

The easiest way to do this is by using synthetic division (I really hope you know what I'm talking about!)

If the factor is x + 2 = 0, then the solution or the root is x = -2.  We will put -2 outside the "box" and the coefficients from the terms of the polynomial inside the "box":

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Bring down the first number and multiply it by the -2.  Put the product up under the first 0 and add:

-2|   1    0    0    8

           -2

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Now multiply the -2 by the -2, put the product up under the next 0 and add:

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Now multiply the 4 by the -2, put the product up under the 8 and add:

-2|   1    0    0    8

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     1    -2   4    0

Those bold numbers are the coefficients to the quotient, which is another polynomial, but is one degree less than the one you started with:

x² - 2x + 4 = 0

Since there is no remainder, then x = -2 is a zero of x³ + 8, and (x + 2) is a factor.

There is a whole world of stuff to learn about quadratics!!

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