Wild Card
What is the excluded value(s)?
5/x
x = 0
(16c^2b^4)/(8c^3b)
(2b^3)/(c); c≠0
(2x^3)/(7x) xx 14/x
4x; x≠0
(x^5)/y div x/y^2
x^4y
(3-2n)/(n(n+2)) - (1-3n)/(n(n+2)
1/n ; n≠0,-2
2/6
Yes; 2 and 6 are both integers and a rational expression is any expression that can be expressed as an integer divided by an integer.
"______, ______, ______" is a method used to divide rational expressions by setting the problem up with multiplication.
KEEP, CHANGE, FLIP
(n+3)/(n^2+10n+21)
1/(n+7); n≠-3, -7
(10n^2)/4 xx 2/n
5n; n≠0
(n^2+7n+12)/(16n^2) div (n+3)/(2n)
(n+4)/(8n); n≠-3,0
c/(c^2+3c-4)-1/(c^2+3c-4)
1/(c+4); c≠-4,1
log(28)
No; log(28) can never be expressed as an integer divided by an integer, because when evaluated it's decimal place never terminates.
What is a rational expression?
A ratio of two polynomials.
Simplify the rational expression
(y^2+9y-10)/(2y+20)
(y-1)/2; y≠-10
(12c^3)/(21b) xx (14b^2)/(6b)
(4bc^2)/3; b≠0
(b+4)/(3b+2) div (3b+12)/(b+1)
(b+1)/((3)(3b+2)); b≠-4,-2/3
(6n)/(n+2)+(2n)/(n+7)
(8n^2+46n)/((n+7)(n+2)); n≠-7,-2
(2sqrtx)/(5x^5)
No; the square root of x could represent a number with a non-terminating decimal. Rational numbers are expressible by a numerator and a denominator that are both integers.
In the expression (x+4)/4 why can't we cancel out the 4's to simplify this expression to equal x?
"4" isn't a factor in the numerator. "4" is only a factor in the denominator. (x+4) is a factor of itself and can not be factored further.
Simplify the rational expression
(x^2-2x-35)/(x^2-9x+14)
(x+5)/(x-2); x≠7, 2
(n^2+n-2)/(n+2) xx (4n)/(n-1)
4n; n≠1, -2
(n^2+14n+48)/(n+1)÷(n+8)
(n+6)/(n+1); n≠-6,-1
(6)/(3x-24)-(7)/(x-7)
(-5x+42)/((x-8)(x-7)); x≠8,-7
x4+360
Yes; this expression can always be expressed by an integer divided by an integer, regardless of the value of x.
What are the excluded value(s)?
(x^2-x-6)/(x^2+8x+12)
x = -6 and -2
Simplify the rational expression
(5d+15)/(d^2-d-12)
5/(d-4); d≠4, -3
(y^2-1)/(y^2-49) xx (y-7)/(y+1)
(y-1)/(y+7); y≠-1, 7,-7
(6k^2)/(6+5k-k^2)÷(k-3)/(k^2-5k-6)
-(6k^2)/(k-3); k≠3, 6,-1
(4)/(3b+1)-(6b)/(4b+4)
(-9b^2+5b+8)/(2(b+1)(3b+1)); n≠-1, -1/3
(-4x^3-3x+5)/(31x^2+4x+600)
Yes; regardless of the value of x the expression will represent an integer divided by another integer.