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Split into factors
Factor out x
Here we have a quadratic trinomial of the form ax^2+bx+c, where |a| ≠1 and there are no common factors. To factor this expression, we will rewrite the middle term, bx, as two terms. The coefficients of these two terms will be factors of ac whose sum must be b. x( 5x^2+13x-6 ) ⇕ x( 5x^2+13x+(- 6) ) We have that a= 5, b=13, and c=- 6. There are now three steps we need to follow in order to rewrite the above expression.
c|c|c|c 1^(st)Factor &2^(nd)Factor &Sum &Result - 1 &30 &- 1 + 30 &29 - 2 & 15 & - 2 + 15 &13 - 3 &10 &- 3 + 10 &7 - 5 &6 &- 5 + 6 &1
Split into factors
Factor out 2x
Here we have a quadratic trinomial of the form ax^2+bx+c, where |a| ≠1 and there are no common factors. To factor this expression we will rewrite the middle term, bx, as two terms. The coefficients of these two terms will be factors of ac whose sum must be b. 2( 3t^2-13t+4 ) ⇕ 2( 3t^2+(- 13)t+4 ) We have that a= 3, b=- 13, and c=4. There are now three steps we need to follow in order to rewrite the above expression.
c|c|c|c 1^(st)Factor &2^(nd)Factor &Sum &Result - 1 & - 12 & - 1 + ( - 12) &- 13 - 2 &- 6 &- 2 + (- 6) &- 8 - 3 &- 4 &- 3 + (- 4) &- 7
The result of factoring out a GCF from the given expression is a binomial with a leading coefficient of 1. 6( x^2-4) Let's temporarily only focus on this binomial, and we will bring back the GCF after factoring.
Do you notice that the binomial is a difference of two perfect squares? This can be factored using the difference of squares method. a^2 - b^2 ⇔ (a+b)(a-b) To do so, we first need to express each term as a perfect square.
Expression | x^2-4 |
---|---|
Rewrite as Perfect Squares | x^2 - 2^2 |
Apply the Formula | (x+2)(x-2) |
Wait! Before we finish, remember that we factored out a GCF from the original expression. To fully complete the factored expression, let's reintroduce that GCF now. 6(x+2)(x-2)