If x³ + 2x and x³ − 7x² + 4x +2 + k have a common linear polynomial factor, then the value of k is what?

Mathematics

1 answer:

Georgia [21]3 years ago

8 0

Answer:

k = -2

Step-by-step explanation:

Factor x³ + 2x:

x³ + 2x = x(x² + 2)

1. The common factor is x, then:

2 + k = 0 ⇒ k = -2 (all the rest are divisible by x)

2. Factor x² + 2 is not linear therefore we discard it.

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Answer:

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3 years ago

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Anyone knows how to do questions 7 and 8? 15 pts!!

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7) Certainly there is a typo in the statement, just see that the expression of item (ii) is different from that of item (i). Probably the correct expression is: $2x^2-4x+5$ . With this consideration, we can continue.

(i) Let E the expression that we are analyzing:

$E=2x^2-4x+5\\\\ E=2x^2-4x+2-2+5\\\\ E=2(x^2-2x+1)-2+5\\\\ E=2(x-1)^2+3$

Since (x-1)² is a perfect square, it is a positive number. So, E is a result of a sum of two positive numbers, 2(x-1)² and 3. Hence, E is a positive number, too.

(ii) Manipulating the expression:

$2x^2+5=4x\\\\ 2x^2-4x+5=0$

So, it's the case when E=0. However, E is always a positive number. Then, there is no real number x that satisfies the expression.

8) Let E the expression that we want to calculate:

$E=(2+1)(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)+1\\\\ E-1=(2+1)(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)$

Multiplying by (2-1) in the both sides:

$(2-1)(E-1)=(2-1)(2+1)(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)\\\\ (E-1)=\underbrace{(2-1)(2+1)}_{2^2-1}(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)\\\\ (E-1)=\underbrace{(2^2-1)(2^2+1)}_{2^4-1}(2^4+1)\cdot ...\cdot(2^{32}+1)\\\\ ...$ Repeating the process, we obtain: $...\\\\ E-1=(2^{32}-1)(2^{32}+1)\\\\ E-1=2^{64}-1\\\\ \boxed{E=2^{64}}$