Thread: Factoring Trinomials

Could anyone help me with this. I know its easy but, I completely forgot everything about math last year. (grade 11 now)

Practice Factoring Trinomials.

determine two values of c so that each expression can be factored.

a) x^2 + 4x + c
b) x^2 - 9x + c

and

Determine expressions to represent the dimensions of this rectangular prism.

Volume is x^3 + 5x^2 + 6x.

please explain ? thanks !

In general,
x^2 + bx + c can be factored into (x + p)(x + q) when:
1) p*q = c, and
2) p + q = b

So for your first question, we require that p + q = 4. So find two numbers with sum = 4 (there will be infinite solutions). For example
p = 104, q = -100 gives c = p*q = -10400
So x^2 + 4x + c = x^2 + 4x - 10400 = (x + 104)(x - 100)

Originally Posted by Thenewguy

Could anyone help me with this. I know its easy but, I completely forgot everything about math last year. (grade 11 now)

Practice Factoring Trinomials.

determine two values of c so that each expression can be factored.

a) x^2 + 4x + c
b) x^2 - 9x + c

and

Determine expressions to represent the dimensions of this rectangular prism.

Volume is x^3 + 5x^2 + 6x.

please explain ? thanks !

Try completing the square for a) and b)



So in order for this quadratic to be factorised over the integers, needs to be a perfect square. What are some values of that will enable this?

ax^2+bx+c can be factored if and only if sqrt(b^2-4ac) is a rational number.
in you above 2 example:
a) x^2 + 4x + c,
It can be factored if c = 4 or 3 or 0..etc...
x^2+4x+4 = (x+2)^2, x^2+4x+3 = (x+3)(x+1), x^2+4x+0 = x(x+4)

b) x^2 - 9x + c
It can be factored if c = -10 or 20 or 0..etc...
x^2-9x-10 = (x-10)(x+1), x^2-9x+20= (x-4)(x-5), x^2-9x+0 = x(x-9)


for the last question, factorize x^3 + 5x^2 + 6x first.
=x^3 + 5x^2 + 6x
=x(x^2+5x+6), you know what to do next

Thank you Piscoau, Prove it, The Chaz. I understand now xD.