Subtract these polynomials. (3x^2+6x+7) - (6x^2-5x-7)

Mathematics

1 answer:

mario62 [17]3 years ago

7 0

Answer:

3x²-11x-14=0

Step-by-step explanation:

Given equation is : (3x²+6x+7)-(6x²-5x-7)

Putting the equation equals to zero making it in the form of polynomial equation (3x²+6x+7)-(6x²-5x-7)=0

Opening the bracket of the equation, the negative sign outside the bracket changes the sign of the values after opening the bracket.

3x²+6x+7-6x²+5x+7=0

Solving this by subtracting the values of same type i.e subtracting the component of x having the same power.

3x²-6x²+6x+5x+7+7=0

-3x²+11x+14=0 OR 3x²-11x-14=0

So, on subtracting the above equation the quadratic equation is obtained 3x²-11x-14=0

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KIM [24]

Answer:

(a) 0

(b) f(x) = g(x)

Step-by-step explanation:

Given rational function:

$f(x)=\dfrac{x^2+2x+1}{x^2-1}$

Part (a)

Factor the numerator and denominator of the given rational function:

$\begin{aligned} \implies f(x) & = \dfrac{x^2+2x+1}{x^2-1} \\\\& = \dfrac{(x+1)^2}{(x+1)(x-1)}\\\\& = \dfrac{x+1}{x-1}\end{aligned}$

Substitute x = -1 to find the limit:

$\displaystyle \lim_{x \to -1}f(x)=\dfrac{-1+1}{-1-1}=\dfrac{0}{-2}=0$

Therefore:

$\displaystyle \lim_{x \to -1}f(x)=0$

Part (b)

From part (a), we can see that the simplified function f(x) is the same as the given function g(x). Therefore, f(x) = g(x).

Part (c)

As x = 1 is approached from the right side of 1, the numerator of the function is positive and approaches 2 whilst the denominator of the function is positive and gets smaller and smaller (approaching zero). Therefore, the quotient approaches infinity.

$\displaystyle \lim_{x \to 1^+} f(x)=\dfrac{\to 2^+}{\to 0^+}=\infty$