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tester [92]
3 years ago
11

Factorise 6x²-14x-12​

Mathematics
1 answer:
maksim [4K]3 years ago
8 0

Answer:

2(x - 3)(3x + 2)

Step-by-step explanation:

Given

6x² - 14x - 12 ← factor out 2 from each term

= 2(3x² - 7x - 6) ← factor the quadratic

Consider the factors of the product of the coefficient of the x² term and the constant term which sum to give the coefficient of the x- term.

product = 3 × - 6 = - 18 and sum = - 7

The factors are - 9 and + 2

Use these factors to split the x- term

3x² - 9x + 2x - 6 ( factor the first/second and third/fourth terms )

= 3x(x - 3) + 2(x - 3) ← factor out (x - 3) from each term

= (x - 3)(3x + 2)

Thus

3x² - 7x - 6 = (x - 3)(3x + 2) and

6x² - 14x - 12 = 2(x - 3)(3x + 2) ← in factored form

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A) Evaluate the limit using the appropriate properties of limits. (If an answer does not exist, enter DNE.)
Gelneren [198K]

For purely rational functions, the general strategy is to compare the degrees of the numerator and denominator.

A)

\displaystyle \lim_{x\to\infty} \frac{2x^2-5}{7x^2+x-3} = \boxed{\frac27}

because both numerator and denominator have the same degree (2), so their end behaviors are similar enough that the ratio of their coefficients determine the limit at infinity.

More precisely, we can divide through the expression uniformly by <em>x</em> ²,

\displaystyle \lim_{x\to\infty} \frac{2x^2-5}{7x^2+x-3} = \lim_{x\to\infty} \frac{2-\dfrac5{x^2}}{7+\dfrac1x-\dfrac3{x^2}}

Then each remaining rational term converges to 0 as <em>x</em> gets arbitrarily large, leaving 2 in the numerator and 7 in the denominator.

B) By the same reasoning,

\displaystyle \lim_{x\to\infty} \frac{5x-3}{2x+1} = \boxed{\frac52}

C) This time, the degree of the denominator exceeds the degree of the numerator, so it grows faster than <em>x</em> - 1. Dividing a number by a larger number makes for a smaller number. This means the limit will be 0:

\displaystyle \lim_{x\to-\infty} \frac{x-1}{x^2+8} = \boxed{0}

More precisely,

\displaystyle \lim_{x\to-\infty} \frac{x-1}{x^2+8} = \lim_{x\to-\infty}\frac{\dfrac1x-\dfrac1{x^2}}{1+\dfrac8{x^2}} = \dfrac01 = 0

D) Looks like this limit should read

\displaystyle \lim_{t\to\infty}\frac{\sqrt{t}+t^2}{3t-t^2}

which is just another case of (A) and (B); the limit would be

\displaystyle \lim_{t\to\infty}\frac{\sqrt{t}+t^2}{3t-t^2} = -1

That is,

\displaystyle \lim_{t\to\infty}\frac{\sqrt{t}+t^2}{3t-t^2} = \lim_{t\to\infty}\frac{\dfrac1{t^{3/2}}+1}{\dfrac3t-1} = \dfrac1{-1} = -1

However, in case you meant something else, such as

\displaystyle \lim_{t\to\infty}\frac{\sqrt{t+t^2}}{3t-t^2}

then the limit would be different:

\displaystyle \lim_{t\to\infty}\frac{\sqrt{t^2}\sqrt{\dfrac1t+1}}{3t-t^2} = \lim_{t\to\infty}\frac{t\sqrt{\dfrac1t+1}}{3t-t^2} = \lim_{t\to\infty}\frac{\sqrt{\dfrac1t+1}}{3-t} = 0

since the degree of the denominator is larger.

One important detail glossed over here is that

\sqrt{t^2} = |t|

for all real <em>t</em>. But since <em>t</em> is approaching *positive* infinity, we have <em>t</em> > 0, for which |<em>t</em> | = <em>t</em>.

E) Similar to (D) - bear in mind this has the same ambiguity I mentioned above, but in this case the limit's value is unaffected -

\displaystyle \lim_{x\to\infty} \frac{x^4}{\sqrt{x^8+9}} = \lim_{x\to\infty}\frac{x^4}{\sqrt{x^8}\sqrt{1+\dfrac9{x^8}}} = \lim_{x\to\infty}\frac{x^4}{x^4\sqrt{1+\dfrac9{x^8}}} = \lim_{x\to\infty}\frac1{\sqrt{1+\dfrac9{x^8}}} = \boxed{1}

Again,

\sqrt{x^8} = |x^4|

but <em>x</em> ⁴ is non-negative for real <em>x</em>.

F) Also somewhat ambiguous:

\displaystyle \lim_{x\to\infty}\frac{\sqrt{x+5x^2}}{3x-1} = \lim_{x\to\infty}\frac{\sqrt{x^2}\sqrt{\dfrac1x+5}}{3x-1} = \lim_{x\to\infty}\frac{x\sqrt{\dfrac1x+5}}{3x-1} = \lim_{x\to\infty}\frac{\sqrt{\dfrac1x+5}}{3-\dfrac1x} = \dfrac{\sqrt5}3

or

\displaystyle \lim_{x\to\infty}\frac{\sqrt{x}+5x^2}{3x-1} = \lim_{x\to\infty}x \cdot \lim_{x\to\infty}\frac{\dfrac1{\sqrt x}+5x}{3x-1} = \lim_{x\to\infty}x \cdot \lim_{x\to\infty}\frac{\dfrac1{x^{3/2}}+5}{3-\dfrac1x} = \frac53\lim_{x\to\infty}x = \infty

G) For a regular polynomial (unless you left out a denominator), the leading term determines the end behavior. In other words, for large <em>x</em>, <em>x</em> ⁴ is much larger than <em>x</em> ², so effectively

\displaystyle \lim_{x\to\infty}(x^4-2x) = \lim_{x\to\infty}x^4 = \boxed{\infty}

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2 years ago
Goods sold to Raman for rs 16000 &amp; received partiat payment of 6000​
lawyer [7]

Step-by-step explanation:

1.RamanA/c dr.6000

to sale a/c 16000

2.cash a/c.6000

to remain 6000

4 0
3 years ago
Martin charges $15 for every 5 bags of leaves he rakes. Last weekend, he raked 26 bags of leaves. How much money did he earn
son4ous [18]
$78 is what he earned.  15/5=3.  3*26=78
7 0
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A random sample of dogs at different animal shelters in a city shoes that 12 of the 60 dogs are puppies
faust18 [17]
1/5th of the dogs are puppies, or 20%, is that what you wanted?

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5. What is the meaning of the statement
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Answer:

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