Math Magic: LCM & GCD Explorer
Discover the secrets of polynomials with our colorful guide!
Find LCM & GCD with Verification
Find the LCM and GCD for the following and verify that f(x) × g(x) = LCM × GCD
GCD is the product of the smallest powers of common factors.
Numerical part: GCD of 21 and 35 is 7
Variable part: GCD of x²y and xy² is xy (lowest power of x is 1, lowest power of y is 1)
∴ GCD = 7xy
LCM is the product of the highest powers of all factors.
Numerical part: LCM of 21 and 35 is 105
Variable part: LCM of x²y and xy² is x²y² (highest power of x is 2, highest power of y is 2)
∴ LCM = 105x²y²
We need to verify that f(x) × g(x) = LCM × GCD
f(x) = 21x²y, g(x) = 35xy²
f(x) × g(x) = 21x²y × 35xy² = 735x³y³
LCM × GCD = 105x²y² × 7xy = 735x³y³
Both sides equal 735x³y³, so the verification is successful!
✅ Verification passed! Our answers are correct.
First polynomial: (x³−1)(x+1)
x³−1 = (x−1)(x²+x+1)
So, f(x) = (x−1)(x²+x+1)(x+1)
Second polynomial: (x³+1)
x³+1 = (x+1)(x²−x+1)
So, g(x) = (x+1)(x²−x+1)
GCD is the product of common factors with lowest power.
Common factor: (x+1)
∴ GCD = (x+1)
LCM is the product of all factors with highest power.
All factors: (x−1), (x²+x+1), (x+1), (x²−x+1)
∴ LCM = (x−1)(x+1)(x²+x+1)(x²−x+1)
f(x) × g(x) = (x−1)(x²+x+1)(x+1) × (x+1)(x²−x+1)
= (x−1)(x²+x+1)(x+1)²(x²−x+1)
LCM × GCD = (x−1)(x+1)(x²+x+1)(x²−x+1) × (x+1)
= (x−1)(x²+x+1)(x+1)²(x²−x+1)
Both sides are equal, so verification is successful!
✅ Verification passed! Our answers are correct.
First polynomial: (x²y + xy²)
= xy(x + y)
Second polynomial: (x² + xy)
= x(x + y)
GCD is the product of common factors with lowest power.
Common factors: x and (x+y)
∴ GCD = x(x + y)
LCM is the product of all factors with highest power.
All factors: x, y, (x+y)
Highest powers: x (from first polynomial), y (from first polynomial), (x+y)
∴ LCM = xy(x + y)
f(x) × g(x) = xy(x + y) × x(x + y) = x²y(x + y)²
LCM × GCD = xy(x + y) × x(x + y) = x²y(x + y)²
Both sides equal x²y(x + y)², so verification is successful!
✅ Verification passed! Our answers are correct.
Find LCM with Given GCD
Find the LCM of each pair of the following polynomials
First polynomial: a² + 4a −12
= a² + 6a - 2a -12 = a(a+6) -2(a+6) = (a-2)(a+6)
Second polynomial: a² −5a + 6
= a² -3a -2a +6 = a(a-3) -2(a-3) = (a-2)(a-3)
We know GCD = (a-2)
f(x) = (a-2)(a+6)
g(x) = (a-2)(a-3)
f(x) × g(x) = (a-2)(a+6) × (a-2)(a-3) = (a-2)²(a+6)(a-3)
Since LCM × GCD = f(x) × g(x)
LCM = (f(x) × g(x)) / GCD = [(a-2)²(a+6)(a-3)] / (a-2) = (a-2)(a+6)(a-3)
LCM = (a-2)(a+6)(a-3)
We can also expand this if needed:
= (a-2)(a² + 3a -18) = a³ + 3a² -18a -2a² -6a +36 = a³ + a² -24a +36
✅ Solution complete! We've found the LCM using the given GCD.
First polynomial: x⁴ -27a³x
= x(x³ -27a³) = x(x-3a)(x²+3a x+9a²)
Second polynomial: (x-3a)²
Already factored.
We know GCD = (x-3a)
f(x) = x(x-3a)(x²+3a x+9a²)
g(x) = (x-3a)²
f(x) × g(x) = x(x-3a)(x²+3a x+9a²) × (x-3a)² = x(x-3a)³(x²+3a x+9a²)
Since LCM × GCD = f(x) × g(x)
LCM = (f(x) × g(x)) / GCD = [x(x-3a)³(x²+3a x+9a²)] / (x-3a) = x(x-3a)²(x²+3a x+9a²)
LCM = x(x-3a)²(x²+3a x+9a²)
✅ Solution complete! We've found the LCM using the given GCD.
Find GCD with Given LCM
Find the GCD of each pair of the following polynomials
First polynomial: 12(x⁴ -x³)
= 12x³(x -1)
Second polynomial: 8(x⁴ −3x³ +2x²)
= 8x²(x² -3x +2) = 8x²(x-1)(x-2)
We know LCM = 24x³(x-1)(x-2)
f(x) = 12x³(x-1)
g(x) = 8x²(x-1)(x-2)
f(x) × g(x) = 12x³(x-1) × 8x²(x-1)(x-2) = 96x⁵(x-1)²(x-2)
Since LCM × GCD = f(x) × g(x)
GCD = (f(x) × g(x)) / LCM = [96x⁵(x-1)²(x-2)] / [24x³(x-1)(x-2)] = 4x²(x-1)
GCD = 4x²(x-1)
✅ Solution complete! We've found the GCD using the given LCM.
First polynomial: (x³ +y³)
= (x+y)(x² -xy +y²)
Second polynomial: (x⁴ +x²y² +y⁴)
Let's add and subtract x²y²:
= x⁴ +2x²y² +y⁴ -x²y² = (x²+y²)² -(xy)² = (x²+y²+xy)(x²+y²-xy)
We know LCM = (x³ +y³)(x² +xy +y²) = (x+y)(x² -xy +y²)(x² +xy +y²)
f(x) = (x+y)(x² -xy +y²)
g(x) = (x²+y²+xy)(x²+y²-xy)
Notice that (x²+y²-xy) is same as (x² -xy +y²)
f(x) × g(x) = (x+y)(x² -xy +y²) × (x²+y²+xy)(x² -xy +y²)
Since LCM × GCD = f(x) × g(x)
GCD = (f(x) × g(x)) / LCM = [(x+y)(x² -xy +y²)²(x²+y²+xy)] / [(x+y)(x² -xy +y²)(x² +xy +y²)]
Simplify: = (x² -xy +y²)(x²+y²+xy) / (x² +xy +y²)
Notice that (x²+y²+xy) is same as (x² +xy +y²), so they cancel out
∴ GCD = (x² -xy +y²)
GCD = (x² -xy +y²)
✅ Solution complete! We've found the GCD using the given LCM.
Find Unknown Polynomial
Given the LCM and GCD of two polynomials p(x) and q(x), find the unknown polynomial
LCM = a³ −10a² +11a +70
GCD = a-7
p(x) = a² −12a +35
We need to find q(x)
p(x) = a² −12a +35
= a² -7a -5a +35 = a(a-7) -5(a-7) = (a-7)(a-5)
q(x) = (LCM × GCD) / p(x)
First, factor LCM:
a³ −10a² +11a +70
Try a=7: 343-490+77+70 = 0 ⇒ (a-7) is a factor
Divide LCM by (a-7):
= (a-7)(a² -3a -10) = (a-7)(a-5)(a+2)
Now calculate:
q(x) = [(a-7)(a-5)(a+2) × (a-7)] / [(a-7)(a-5)] = (a-7)(a+2) = a² -5a -14
q(x) = a² -5a -14
✅ Solution complete! We've found the unknown polynomial q(x).
LCM = (x⁴ −y⁴)(x⁴ +x²y² +y⁴)
GCD = (x² -y²)
q(x) = (x⁴ −y⁴)(x² +y² −xy)
We need to find p(x)
First, factor LCM:
(x⁴ −y⁴) = (x²-y²)(x²+y²) = (x-y)(x+y)(x²+y²)
(x⁴ +x²y² +y⁴) = (x²+y²)² -x²y² = (x²+y²+xy)(x²+y²-xy)
So LCM = (x-y)(x+y)(x²+y²)(x²+y²+xy)(x²+y²-xy)
GCD = (x² -y²) = (x-y)(x+y)
q(x) = (x⁴ −y⁴)(x² +y² −xy) = (x-y)(x+y)(x²+y²)(x²+y²-xy)
p(x) = (LCM × GCD) / q(x)
Substitute the factored forms:
Numerator: LCM × GCD = [(x-y)(x+y)(x²+y²)(x²+y²+xy)(x²+y²-xy)] × [(x-y)(x+y)]
= (x-y)²(x+y)²(x²+y²)(x²+y²+xy)(x²+y²-xy)
Denominator: q(x) = (x-y)(x+y)(x²+y²)(x²+y²-xy)
Divide numerator by denominator:
p(x) = (x-y)(x+y)(x²+y²+xy)
p(x) = (x-y)(x+y)(x²+y²+xy)
We can also write this as:
p(x) = (x²-y²)(x²+y²+xy)
✅ Solution complete! We've found the unknown polynomial p(x).