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topjm [15]
3 years ago
13

a polynomial has one quadratic factor and 3 linear factors. One of the linear factors has multiplicity two. What is the degree o

f the polynomial
Mathematics
1 answer:
snow_lady [41]3 years ago
8 0

Answer:

5

Step-by-step explanation:

a polynomial has one quadratic factor and 3 linear factors. One of the linear factors has multiplicity two. What is the degree of the polynomial

A polynomial with one quadratic obtains the forms ( ax² +bx +c ) with 3 linear factors.

Suppose the three linear fractions are :

(x- P) (x-Q) (x- R)

∴

The polynomial = ( ax² +bx +c )(x- P) (x-Q) (x- R)

By factorization, the highest degree of the polynomial = 5

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Aleksandr-060686 [28]

Answer:

case a) x^{2}=3y ----> open up

case b) x^{2}=-10y ----> open down

case c) y^{2}=-2x ----> open left

case d) y^{2}=6x ----> open right

Step-by-step explanation:

we know that

1) The general equation of a vertical parabola is equal to

y=a(x-h)^{2}+k

where

a is a coefficient

(h,k) is the vertex

If a>0 ----> the parabola open upward and the vertex is a minimum

If a<0 ----> the parabola open downward and the vertex is a maximum

2) The general equation of a horizontal parabola is equal to

x=a(y-k)^{2}+h

where

a is a coefficient

(h,k) is the vertex

If a>0 ----> the parabola open to the right

If a<0 ----> the parabola open to the left

Verify each case

case a) we have

x^{2}=3y

so

y=(1/3)x^{2}

a=(1/3)

so

a>0

therefore

The parabola open up

case b) we have

x^{2}=-10y

so

y=-(1/10)x^{2}

a=-(1/10)

a

therefore

The parabola open down

case c) we have

y^{2}=-2x

so

x=-(1/2)y^{2}

a=-(1/2)

a

therefore

The parabola open to the left

case d) we have

y^{2}=6x

so

x=(1/6)y^{2}

a=(1/6)

a>0

therefore

The parabola open to the right

3 0
2 years ago
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