Explanation:
If #a# is a zero of a polynomial in #x#, then #(x-a)# is a factor and vice versa.
So a polynomial of minimum degree with zeros#-1#, #1# and #5# is:
#(x+1)(x-1)(x-5) = (x^2-1)(x-5) = x^3-5x^2-x+5#
and a polynomial equation of minimum degree with roots #-1#, #1# and #5# is:
#x^3-5x^2-x+5 = 0#
Any non-zero constant multiple of this cubic equation is also a soltuion.
Explanation:
#"given the roots of a polynomial are"#
#x=a,x=b,x=c,x=d#
#"then the factors of the polynomial are"#
#(x-a),(x-b),(x-c)" and "(x-d)#
#"the polynomial is the the product of the factors"#
#p(x)=(x-a)(x-b)(x-c)(x-d)#
#"here "x=-2,x=-0.5,x=4" are the roots"#
#rArr(x+2),(x+0.5)" and "(x-4)" are the factors"#
#rArrp(x)=(x+2)(x+0.5)(x-4)#
#color(white)(rArrp(x))=2x^3-3x^2-18x-8" possible polynomial"#
Sally H. Write a polynomial of least degree with roots 1, 5, and -6.
Write your answer using the variable x and in standard form with a leading coefficient of 1.
3 Answers By Expert Tutors
Isidro L. answered • 08/14/19
AP Calculus AB Teacher 20 years Experience.
The process for answer this question is very easy if you remember the property of zero for a polynomial function, meaning that if (x-2)=0, and you solve for x , x-2=0, adding 2 in both sides we get x=2.
In our case the problem provides the x values, x=1, x=5 and x=-6.
Therefore : P(x) = (x-1) (x-5)(x+6). Find the products of these factors(FOIL)
2) f(x)= (x^2-5x-x+5)(x+6)=(x^2-6x+5) (x+6)
3) f(x)=(x^2-6x+5)(x+6)=(x^3+6x^2-6x^2-36x+5x+30). So your answer should be F(x)=x^3-31x+30
Note: To verify your answer , graph the function in a TI-84 or demos calculator (free online) and verify the interceptions. in this case you notice that graph passing trough x=1, y=0, x=5,y=0 and x=-6, y=0.
Dick S. answered • 08/14/19
Science Tech Eng Math Physics, Simple Explanations, UC Berkeley Grad
When you factor a polynomial you end up with it's roots.
To find it's roots you set it equal to zero and solve for all
possible values of x.
for example, a polynomial of degree two, or second order
polynomial, (2 being the highest power of x present
in the polynomial) in the form:
ax2 + bx + c = 0
has two roots,
x1, x2 = [-b ± √(b2 - 4ac)] / 2a
which factors the above polynomial into factors:
ax2 + bx + c = 0 = (x - x1)(x - x2)
If we have three roots we have polynomial of at least degree
and it can be factored into 3 roots:
(x - x1)(x - x2)(x - x3)
in this case we have
x1 = 1, x2 = 5, x3 = -6
which gives a polynomial that is the product of at least
the following three factors:
(x - 1)(x - 5)(x + 6)
which, when multiplied out yields the polynomial:
(x - 1)(x - 5)(x + 6) = (x2 - x - 5x + 5)(x + 6) = (x2 - 6x + 5)(x + 6)
= x3 - 6x2 + 5x + 6x2 - 36x + 30
= x3 - 31x + 30 answer
If you are given the roots, you can find the factors and then multiply the factors together to get the polynomial. For the roots provided, the factors would be:
f(x) = (x-1)(x-5)(x+6) (The signs are opposite because if you set these factors equal to zero and solve, you will get the roots provided)
Now multiply. It doesn't matter which terms you do in which order. You can start with FOIL for the first set of binomials.
f(x) = (x2-x-5x+5)(x+6) Combine like terms.
f(x) = (x2-6x+5)(x+6) Now multiply this using the distributive property distributing (x+6) to each term
f(x) = (x+6) (x2)+(x+6)(-6x)+(x+6)(5) Now distribute again.
f(x) = (x2)(x)+(x2)(6)+(-6x)(x)+(-6x)(6)+(5)(x)+(5)(6) Multiply
f(x) = x3+6x2-6x2-36x+5x+30 Combine like terms
f(x) = x3-31x+30
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