Answer:
{x}^{4}+4{x}^{3}y+6{x}^{2}{y}^{2}+4x{y}^{3}+{y}^{4}x
4
+4x
3
y+6x
2
y
2
+4xy
3
+y
4
Step-by-step explanation:
1 Use Square of Sum: {(a+b)}^{2}={a}^{2}+2ab+{b}^{2}(a+b)
2
=a
2
+2ab+b
2
.
({x}^{2}+2xy+{y}^{2})({x}^{2}+2xy+{y}^{2})(x
2
+2xy+y
2
)(x
2
+2xy+y
2
)
2 Expand by distributing sum groups.
{x}^{2}({x}^{2}+2xy+{y}^{2})+2xy({x}^{2}+2xy+{y}^{2})+{y}^{2}({x}^{2}+2xy+{y}^{2})x
2
(x
2
+2xy+y
2
)+2xy(x
2
+2xy+y
2
)+y
2
(x
2
+2xy+y
2
)
3 Expand by distributing terms.
{x}^{4}+2{x}^{3}y+{x}^{2}{y}^{2}+2xy({x}^{2}+2xy+{y}^{2})+{y}^{2}({x}^{2}+2xy+{y}^{2})x
4
+2x
3
y+x
2
y
2
+2xy(x
2
+2xy+y
2
)+y
2
(x
2
+2xy+y
2
)
4 Expand by distributing terms.
{x}^{4}+2{x}^{3}y+{x}^{2}{y}^{2}+2{x}^{3}y+4{x}^{2}{y}^{2}+2x{y}^{3}+{y}^{2}({x}^{2}+2xy+{y}^{2})x
4
+2x
3
y+x
2
y
2
+2x
3
y+4x
2
y
2
+2xy
3
+y
2
(x
2
+2xy+y
2
)
5 Expand by distributing terms.
{x}^{4}+2{x}^{3}y+{x}^{2}{y}^{2}+2{x}^{3}y+4{x}^{2}{y}^{2}+2x{y}^{3}+{y}^{2}{x}^{2}+2{y}^{3}x+{y}^{4}x
4
+2x
3
y+x
2
y
2
+2x
3
y+4x
2
y
2
+2xy
3
+y
2
x
2
+2y
3
x+y
4
6 Collect like terms.
{x}^{4}+(2{x}^{3}y+2{x}^{3}y)+({x}^{2}{y}^{2}+4{x}^{2}{y}^{2}+{x}^{2}{y}^{2})+(2x{y}^{3}+2x{y}^{3})+{y}^{4}x
4
+(2x
3
y+2x
3
y)+(x
2
y
2
+4x
2
y
2
+x
2
y
2
)+(2xy
3
+2xy
3
)+y
4
7 Simplify.
{x}^{4}+4{x}^{3}y+6{x}^{2}{y}^{2}+4x{y}^{3}+{y}^{4}x
4
+4x
3
y+6x
2
y
2
+4xy
3
+y
4
Answer:
11.5 in
Step-by-step explanation:
Answer:
see explanation
Step-by-step explanation:
The Remainder theorem states that if f(x) is divided by (x - h) then
f(h) is the remainder, thus
division by (x - 1) then h = 1
f(1) = 4(1)³ - 7(1)² - 2(1) + 6
= 4 - 7 - 2 + 6 = 1 ← remainder
The factor theorem states that if (x - h) is a factor of f(x), then f(h) = 0
Here f(1) = 1
Hence (x - 1) is not a factor of f(x)
Answer:
Please don't cheat on your map test. These are very important for your school, I know that lots of kids dont like school but I hate to be a party pooper. You can reportthis answer since its not a answer.
Step-by-step explanation:
Answer:
a = 1
b = -1
c = -2
Step-by-step explanation:
We reorder the equation in such way that let us see the usual
ax² + bx + c = 0
Then the original quation is:
-2 = - x + x² - 4 ⇒ 0 = 2 - x + x² - 4 ⇒ x² - x - 2 = 0
Now we are able by simply inspection to identify a, b . c comparing our equation with the general equation so :
a = 1
b = -1
c = -2
