# 1) Find the zero of the polynomial, P(x)=cx+c

## Answers

0 = x⁴ (x-1)² (x+7)

x= 0

x= 1

x = -7

0.4, ±1

Step-by-step explanation:

Hey there times is 0 when 1 on () is 0 (sorry for language as i am dont speak english that well)

5x-2=0=>x=0.4

X²-1=0=>x=±1

Have a nice day ;)

a) -2, 1 and 2

b) 4

Step-by-step explanation:

The given equation is:

g(x) = (x + 2)(x - 1)(x - 2)

Part a1 and a2) Finding zeros of the function

By zeroes of the function we mean the points the graph where the value of the function is zero. In order to find the zeroes of the function we equate the function equation to zero and find the corresponding values of x. This is shown below:

0 = (x + 2)(x - 1)(x - 2)

According to the zero product property, we can write:

x + 2 = 0 ⇒ x = -2

x - 1 = 0 ⇒ x = 1

x - 2 = 0 ⇒ x = 2

Therefore, the zeroes of the given function are -2, 1 and 2. In ordered pairs we can write these as (-2, 0), (1, 0) and (2, 0)

Part b1 and b2) y-intercept of the function

y-intercept is the point on the graph where the function crosses the y-axis. Since on y-axis, the value of x coordinate is zero, in order to find the y-intercept we simply substitute zero for all occurrences of x. This is shown below:

g(0) = (0 + 2)(0 - 1)(0 - 2)

g(0) = 2(-1)(-2)

g(0) = 4

Thus, the y-intercept of the function is 4, which in ordered pair can be written as (0, 4)

Part c) End behavior of the graph

End behavior of the graph depends on:

Sign of the leading coefficientDegree of the functionDegree of the function is 3 i.e. odd and sign of leading coefficient is positive. This means the end behavior of the graph will be:

Rising towards Right endFalling towards Left endPart d) Graph of polynomial

Based on all the above discussion, a graph of the function can be plotted. The graph is attached below.

[tex]Four Questions, 60 points & Brainliest! (Please show all steps)g(x) = (x + 2)(x - 1)(x - 2)a1) F[/tex]

Answers

1) 4

2) 0, 2, -10

3) 0, 5, -2

Step-by-step explanation:

1) x³ - 64 = 0

x³ = 64

x = 4

2) x³(x² + 8x - 20) = 0

x³(x² + 10x - 2x - 20) = 0

x³(x(x + 10) - 2(x + 10)) = 0

x³(x + 10)(x - 2) = 0

x = 0, 2, -10

3) x³ - 3x² - 10x = 0

x(x² - 3x - 10) = 0

x(x² - 5x + 2x - 10) = 0

x(x(x - 5) + 2(x - 5)) = 0

x(x - 5)(x + 2) = 0

x = 0, 5, -2

See below .

Step-by-step explanation:

This is a cubic function whose graph will contain 2 loops.

a) (x + 2)(x - 1)(x - 2) = 0

Since this a product of 3 terms and the result is zero all 3 could be zero so we have:

x + 2 = 0, x - 1 = 0 or x - 2 = 0

So the zeroes are -2, 1 and 2.

b) the y -intercept occurs when x = 0 , so substituting for x:

y-intercept = (0 + 2)(0 - 1)(0 - 2) = 2*-1*-2 = 4 = y coordinate of the y-intercept

y-intercept is at (0, 4).

c) Plot the points (-2, 0), ( 1, 0), ( 2, 0) and (0, 4).

Also I suggest you you plot some points (-1, 6) and (1.5 , -0.9) where the graph will have loops.

d) As the leading coefficient is positive the graph will fall to the left and rise to the right.

The

Q1: a binomial of degree 100:

for example: x^100 + 2 (it is called binomial because the number of term is 2: x^100 and 2)

Q2: the coefficients:

the main rule is:

if we have P(x)= a1X^p + a2X^p-1+ + a0 as polynomial, so the coefficients are (a1, a2, ... ao)

i) x3 in 2x + x2 – 5x3 + x4, the coefficient is -5

ii) x2 in (π/3) x2 + 7x – 3, the coefficient is (π/3)

iii) x in √3 - 2√2x + 4x2, the coefficient is - 2√2

Q3. If p(y) = 4 + 3y – y2 + 5y3, (i) p(0) can be found with

(i) p(0) = 4 + 3*0 – 0² + 5*0^3= 4, p(0)= 4, the same method to p(2)

ii) p(2)=4 + 3*2– 2² + 5*2^3=49, p(2)=49

iii) p(-1)= 4 + 3*(-1) – (-1)² + 5*(-1)^3 = -5, p(-1)= -5

Q4. the zero of the polynomial

i) h(x) = 6x-1, to find it, just egalize h(x) to 0, 6x-1 =0, it implies 6x=1, and x=1/6, so the zero is x=1/6

ii) p(t) = 2t-3, ii) p(t) = 2t-3=0 implies 2t=3 and the zero is t=3/2

the remainder in each case if p(x) is divided by g(x)

(i) p(x) = x3 – 6x2 + 2x – 4, g(x) = 1 – 2x

x3 – 6x2 + 2x – 4

R(x) = = (-x²/2 + 11x/4 + 3/8) - 35 / 8(-2x + 1)

1 – 2x

so the remainder is - 35 / 8

(ii) p(x) = 2x3 – 5x2 + 4x – 3, g(x) = 3x + 1.

2x3 – 5x2 + 4x – 3

R(x) = = 2x²/3 -17x /9 +53/27 - 134 /27(3x+1)

3x + 1

the remainder is - 134 /27

Read more on -

Step-by-step explanation:

The answer is given below

Step-by-step explanation:

A number is said to be a zero of a polynomial if when the number is substituted into the function the result is zero. That is if a is a zero of polynomial f(x), therefore f(a) = 0.

Since P(−1)=0 P(0)=1 P(2+√3)=0, therefore -1 and 2+√3 are zeros of the polynomial.

Gary is right because there are 2 known zeros of P(x) which are −1 and 2+√3. Also 2 - √3 is also a root. From irrational root theorem, irrational roots are in conjugate pairs i.e. if a+√b is a root, a-√b is also a root.

Heather is not correct because if P(0) = 1, it means that 0 is not a root. It does not mean that 1 is a zero of P(x)

Irene is correct. since P(−1) and P(2+3–√) equal 0, 2 zeros of P(x) are −1 and 2+√3. They may be other zeros of P(x), but there isn't enough information to determine any other zeros of P(x)

Each of the parentheses should be set to zero to find both zeroes.

You end up with 2/5 and 1/2

The

Q1: a binomial of degree 100:

for example: x^100 + 2 (it is called binomial because the number of term is 2: x^100 and 2)

Q2: the coefficients:

the main rule is:

if we have P(x)= a1X^p + a2X^p-1+ + a0 as polynomial, so the coefficients are (a1, a2, ... ao)

i) x3 in 2x + x2 – 5x3 + x4, the coefficient is -5

ii) x2 in (π/3) x2 + 7x – 3, the coefficient is (π/3)

iii) x in √3 - 2√2x + 4x2, the coefficient is - 2√2

Q3. If p(y) = 4 + 3y – y2 + 5y3, (i) p(0) can be found with

(i) p(0) = 4 + 3*0 – 0² + 5*0^3= 4, p(0)= 4, the same method to p(2)

ii) p(2)=4 + 3*2– 2² + 5*2^3=49, p(2)=49

iii) p(-1)= 4 + 3*(-1) – (-1)² + 5*(-1)^3 = -5, p(-1)= -5

Q4. the zero of the polynomial

i) h(x) = 6x-1, to find it, just egalize h(x) to 0, 6x-1 =0, it implies 6x=1, and x=1/6, so the zero is x=1/6

ii) p(t) = 2t-3, ii) p(t) = 2t-3=0 implies 2t=3 and the zero is t=3/2

the remainder in each case if p(x) is divided by g(x)

(i) p(x) = x3 – 6x2 + 2x – 4, g(x) = 1 – 2x

x3 – 6x2 + 2x – 4

R(x) = = (-x²/2 + 11x/4 + 3/8) - 35 / 8(-2x + 1)

1 – 2x

so the remainder is - 35 / 8

(ii) p(x) = 2x3 – 5x2 + 4x – 3, g(x) = 3x + 1.

2x3 – 5x2 + 4x – 3

R(x) = = 2x²/3 -17x /9 +53/27 - 134 /27(3x+1)

3x + 1

the remainder is - 134 /27

Zeros:

You get the zeros through the zero product property so

X+2=0. X-1=0. X-2=0. So your zeros are

-2, 1, and 2.

Y intercept:

To find your y intercept just substitute 0 for x

G(0)=(0+2)(0-1)(0-2) and your y int is 4.

Graphing:

To graph it Plot your zeros first at the right coordinates. So (-2,0), (1,0), and (2,0) since the degree of the zeros are odd the line goes through the coordinates.

Once you get this you have to see how your graphs moves on the left and right.

So you essentially substitute -∞ and ∞ for your x.

Left of graph

G(-∞)= (-)(-)(-) and since you have three negative values your y will also be negatives so it goes down on the left.

Right of graph

G(∞)=(+)(+)(+) since this has three positives values your y will be positive so it goes up on right.