Factoring algebraic expressions
- Thread starter aiwil123
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pka
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What you posted is far too vague for us to help.
Please give us something that tell us what help you need.
There is quite a lot on that attachment. So say what part(s) with which we can help you.
Otis
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Hello aiwil.
11·x2 +88·x +176 = 0
Each of those numbers is a multiple of 11.
11·1·x2 +11·8·x +11·16 = 0
In other words, 11 is a common factor in each term. That means we can "pull out" the common factor, and write it in front of what's left.
11(x2 +8x +16) = 0
The next step is to factor the quadratic polynomial (x2+8x+16).
When we try to factor a polynomial in the form Ax2+Bx+C, we firstr look for two numbers (I'll call them 'number1' and 'number2') whose product is C and whose sum is B. That is, we're hoping to find two numbers so that
number1 × number2 = 16
number1 + number2 = 8
The polynomial's factorization then takes the form
Ax2 + Bx + C = (x + number1)(x + number2)
With a little thought, we realize the two numbers are 4 and 4 because
4 × 4 = 16
4 + 4 = 8
Therefore, the quadratic polynomial factors as (x+4)(x+4), and we can write that as (x+4)2.
(x + 4)2 equals zero when x+4 is zero.
x + 4 = 0
Subtract 4 from each side.
x + 4 - 4 = 0 - 4
x = -4
Questions?
?
11·x2 +88·x +176 = 0
Each of those numbers is a multiple of 11.
11·1·x2 +11·8·x +11·16 = 0
In other words, 11 is a common factor in each term. That means we can "pull out" the common factor, and write it in front of what's left.
11(x2 +8x +16) = 0
The next step is to factor the quadratic polynomial (x2+8x+16).
When we try to factor a polynomial in the form Ax2+Bx+C, we firstr look for two numbers (I'll call them 'number1' and 'number2') whose product is C and whose sum is B. That is, we're hoping to find two numbers so that
number1 × number2 = 16
number1 + number2 = 8
The polynomial's factorization then takes the form
Ax2 + Bx + C = (x + number1)(x + number2)
With a little thought, we realize the two numbers are 4 and 4 because
4 × 4 = 16
4 + 4 = 8
Therefore, the quadratic polynomial factors as (x+4)(x+4), and we can write that as (x+4)2.
(x + 4)2 equals zero when x+4 is zero.
x + 4 = 0
Subtract 4 from each side.
x + 4 - 4 = 0 - 4
x = -4
Questions?
?
I would explain it a bit differently from otis, but fundamentally it is the same idea in a different order.
[MATH]11x^2 + 88x + 176 = 0 \implies \\ 11(x^2) + 8(11x) + 11(16) = 0 \implies \\ \dfrac{1}{11} * \{11(x^2) + 8(11x) + 11(16)\} = \dfrac{1}{11} * 0 \implies\\ x^2 + 8x + 16 = 0 \implies \\ x^2 + 2(4 * x) + 4^2 = 0 \implies \\ (x + 4)^2 = 0 \implies \\ x + 4 = \pm \sqrt{0} = 0 \implies \\ x = -4.[/MATH]
[MATH]11x^2 + 88x + 176 = 0 \implies \\ 11(x^2) + 8(11x) + 11(16) = 0 \implies \\ \dfrac{1}{11} * \{11(x^2) + 8(11x) + 11(16)\} = \dfrac{1}{11} * 0 \implies\\ x^2 + 8x + 16 = 0 \implies \\ x^2 + 2(4 * x) + 4^2 = 0 \implies \\ (x + 4)^2 = 0 \implies \\ x + 4 = \pm \sqrt{0} = 0 \implies \\ x = -4.[/MATH]
No, I get it now! Thank you so much!
You help me too! Thank you!!