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3 years ago

How many values of x must be excluded in the expression x-2/(x+9)(x-5)

2 answers:

5 0

We have the expression:( x - 2 ) / ( x + 9 ) ( x - 5 )We have to find how many values of x must be excluded. The rule is : denominator must never equal zero. Anything divided by zero is undefined.x + 9 ≠ 0, x ≠ - 9x - 5 ≠ 0, x ≠ 5.Answer: Two values must be excluded, x = - 9 and x = 5.

jeka57 [31]3 years ago

5 0

Answer: There is one excluded value and the excluded value is negative four.

Step-by-step explanation:

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3 years ago

F(-12) if f(x) = 2x + 15

Alenkinab [10]

Answer: $f(-12)=-9$

Step-by-step explanation:

We have the following function:

$f(x)=2x+15$

And we have to evaluate it when $x=-12$ , this means we have to substitute $x$ by $-12$ :

$f(-12)=2(-12)+15$

Solving:

$f(-12)=-24+15$

Finally:

$f(-12)=-9$

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4 years ago

Find all the zeros of the polynomial function p(x) = x3 – 5x2 + 33x – 29

hram777 [196]

Answer:

$\large \boxed{\sf \ \ x=1, \ \ x=2+5i, \ \ x=2-5i \ \ }$

Step-by-step explanation:

Hello,

I assume that we are working in $\mathbb{C}$ , otherwise there is only one zero which is 1. Please consider the following.

First of all, <u>we can notice that 1 is a trivial solution</u> as

$p(1) = 1^3-5\cdot 1^2 + 33\cdot 1-29=1-5+33-29=0$

It means that (x-1) is a factor of p(x) so we can find two real numbers, a and b, so that we can write the following.

$p(x)=(x-1)(x^2+ax+b)=x^3+ax^2+bx-x^2-ax-b=x^3+(a-1)x^2+(b-a)x-b$

Let's identify like terms as below.

a-1 = -5 <=> a = -5 + 1 = -4

b-a = 33

-b = -29 <=> b = 29

$\boxed{ \ p(x)=(x-1)(x^2-4x+29) \ }$

Now, we need to find the zeroes of the second factor, meaning finding x so that:

$x^2-4x+29=0 \ \text{ complete the square, 29 = 25 + 4} \\ \\ x^2-2\cdot 2 \cdot x+2^2+25=0 \\ \\ (x-2)^2=-25=(5i)^2 \ \text{ take the root } \\ \\x-2=\pm 5i \ \text{ add 2 } \\ \\ x = 2+5i \ \text{ or } \ x = 2-5i$