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3 years ago

8pq + 20qr – 16s factorise plz

Mathematics

1 answer:

Dafna1 [17]3 years ago

8 0

Answer:

4(2qp + 5qr - 4s)

Step-by-step explanation:

8pq + 20qr – 16s

we can use 4 as a common factor between these numbers.

8/4 = 2

20/4 = 5

16/4 = 4

4(2qp + 5qr - 4s) is the most factorized answer we can get

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2 years ago

What are the real and complex solutions of the polynomial equation? x^4-41x^2=-400. *work needed*

pickupchik [31]

<h3>Answer: x = -5, -4, 4 and 5.</h3><h3>All four zeros are real solutions.</h3>

Step-by-step explanation:

Given the polynomial equation $x^4-41x^2=-400$ .

Adding 400 on both sides to get rid 400 from right side and set 0 on right side, we get

$x^4-41x^2+400=-400+400$ .

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Factoring by product sum rule.

We need product of 400 and sum upto -41.

We can see that 400 = -25 × -16 = 400 and -25-16 = -41.

Therefore,

$x^4-25x^2-16x^2+400=0$

Making it into two groups, we get

$(x^4-25x^2)+(-16x^2+400)=0$

Factoring out GCF of each group, we get

$x^2(x^2-25)-16(x^2-25)=0$

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Factoring out $(x^2-25)$ and $(x^2-16)$ separately by difference of the squares identity $a^2-b^2=(a-b)(a+b)$ , we get

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x-5=0

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x+4=0.

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Step-by-step explanation:

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