Hey, I have a question, you can factor using guess and check or grouping, here it is
-13a + 12 - 4a^2
the answer is -(a+4)(4a-3)
Not sure how to get it using either one, please help! Thanks!
First re-arrange:
$\displaystyle -4a^2 - 13a + 12 $
Now factor out the negative sign
$\displaystyle -(4a^2 + 13a - 12) $
Now we should take the $\displaystyle 4a^2 $ term and factorise that, keeping in mind that it is not yet complete. (The $\displaystyle x_{1} $ and $\displaystyle x_{2} $ terms represent the constants)
$\displaystyle -(4a + x_{1})(a + x_{2}) $
Now we should see which constants give us a product of 12.
(1 x 12 ; 2 x 6 ; 3 x 4)
Now from those options we should find which also satisfies the middle term.
Can you take it from here? Just do it step by step.
2)$\displaystyle n^2(n - 1) + 1(n - 1)$
$\displaystyle = (n^2 + 1)(n - 1)$
3) $\displaystyle (4a - 8)(a + 2)$
4) $\displaystyle -2(3x^3 - 5x^2 + 2x^4)$
$\displaystyle = -2x^2(3x - 5 + 2x^2)$
$\displaystyle = -2x(2x^2 + 3x - 5)$
$\displaystyle = -2x(2x + 5)(x - 1)$
5) $\displaystyle 6x^2 + x - 1$
We know that this value is equal to 2 values multiplied by each other. So let's factorise it:
$\displaystyle (3x - 1)(2x + 1)$
and the answers in the back of the book for the problems you answered are a little different, thanks for helping me, i'm just confused!
2.) n^3-n^2+n-1
(n-1) (n^2 +1)
3.) 4a(a+2) - 8 (a+2)
4(a+2)(a-2)
4.) -6x^3 +10x^2 - 4x^4
-2x^2(2x+5)(x-1)
5.) a rectangle has area described by the polynomial 6x^2 + x -1 suare centimeters. Find the length and width of the rectangle.
2x-5