# Factor the polynomial completely x^3-3x^2-10x?

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Can you show steps please? Thanks!

• Factor: x^3 – 3x^2 – 10

x(x^2 – 3x – 10) = 0

x[(x^2 – 3x + 9/4) – 10 – 9/4] = 0

x[(x – 3/2)^2 – 49/4] = 0

I like to always use completing the square for all quadratics

first x:

x = 0

second x:

[(x – 3/2)^2 – 49/4] = 0

(x – 3/2)^2 – 49/4 = 0

(x – 3/2)^2 = 49/4

x – 3/2 = +/- sqrt (49/4)

x = 3/2 +/- sqrt (49/4)

x = 3/2 +/- 7/2

x = 5 or x = -2

So all your x as factors: (x)(x-5)(x+2)

• 1.) Notice that an x appears in every term. You can factor out this x to make the new form:

x(x^2-3x-10)

2.) Now factor the section inside the parenthesis:

x(x-5)(x+2) <— You do this by noticing that -5 and 2 multiply to make -10 and add to make -3.

• x^(3)-3x^(2)-10x

Factor out the GCF of x from each term in the polynomial.

x(x^(2))+x(-3x)+x(-10)

Factor out the GCF of x from x^(3)-3x^(2)-10x.

x(x^(2)-3x-10)

For a polynomial of the form x^(2)+bx+c, find two factors of c (-10) that add up to b (-3). In this problem 2*-5=-10 and 2-5=-3, so insert 2 as the right hand term of one factor and -5 as the right-hand term of the other factor.

x(x+2)(x-5)

• Its actually quite simple, all you do is factor out the x so the equation would be x(X^2-3x-10), solve the inside as you normally would with a quadratic and then multiply x to the answers.

Work:

x(x^2-3x-10)

x((x-5)(x+2))

Source(s): Algebra 2 Trigonometry as a Sophomore in High School
• x^3 – 3x^2 – 10x

x (x^2 – 3x – 10)

x (x – 5)(x + 2)

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