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yawa3891 [41]
1 year ago
13

8k^4 -3k^2 multiplied by 5k^2

Mathematics
2 answers:
S_A_V [24]1 year ago
6 0

Answer: 40k⁶-15k⁴

Step-by-step explanation:

(8k^4-3k^2)(5k^2)=\\\\8k^4(5k^2)-3k^2(5k^2)=\\\\8(5)(k^{4+2})-3(5)(k^{2+2})=\\\\40k^6-15k^4

jekas [21]1 year ago
3 0

Result of 8k^{4} -3k^{2} multiplied by 5k^{2} is 40k^{6}-15 k^{4}

Algebra and Exponent

To answer this question you need to know about algebra and exponent rule.

8k^{4} -3k^{2} multiplied by 5k^{2}

It means the operation will become:

5k^{2} (8k^{4} -3k^{2} )=

  • First, 8k^{4} -3k^{2} can't be calculate because of different variable. Even though they both use the letters 'k', k^{4} dan k^{2} cannot be added because they have different powers. In algebraic addition, addition can only be done on the same variables and which have the same power.
  • Because 8k^{4} -3k^{2} can not be calculated, then we directly calculate the multiplication 5k^{2} (8k^{4} -3k^{2} )

The way to calculate algebraic multiplication with 2 or more variables is the way I call 'rainbow'.

5k^{2} (8k^{4} -3k^{2} )=

5k^{2} multiplied by 8k^{4},  5k^{2} multiplied by -3k^{2} first, then the two are added.

  • Let's multiply one by one first

(5k^{2}) (8k^{4} )=

The way to multiply this is by: numbers multiplied by numbers and letters multiplied by letters

=5k^{2} (8k^{4} -3k^{2} )\\=((5)(8))((k^{2} )(k^{4} ))\\=40((k^{2} )(k^{4} )

Then how to multiply the letters, this uses the rule of exponents. If there are 2 same variables are multiplied, then the result is a variable with the power of the sum of the two powers.

a^{m} a^{n} =a^{m+n}

Which means,

k^{2}k^{4}  =k^{6}

Then,

=5k^{2} (8k^{4} -3k^{2} )\\=((5)(8))((k^{2} )(k^{4} ))\\=40((k^{2} )(k^{4} )\\=40k^{6}

Also same with above,

=(5k^{2} )(-3k^{2} )\\=((5)(-3))((k^{2} )(k^{2} )\\=-15k^{4}

  • Add the result of two operation above

=5k^{2} (8k^{4} -3k^{2} )\\=((5k^{2})(8k^{4})+((5k^{2})(-3k^{2})\\=(40k^{6})+(-15k^{4} )\\=40k^{6}-15k^{4}

Learn more about algebra multiplication here: https://brainly.ph/question/72890

#SPJ1

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