To solve the expression [tex]\(4a^2 + 4a - 3\)[/tex] through factoring, we need to find two binomial expressions such that their product gives the original quadratic expression.
The general form for factoring a quadratic [tex]\(ax^2 + bx + c\)[/tex] is given by:
[tex]\[
(ax + p)(qx + r)
\][/tex]
Such that when expanded:
[tex]\[
(ax + p)(qx + r) = aqx^2 + (ap + r) x + pr
\][/tex]
For our quadratic expression [tex]\(4a^2 + 4a - 3\)[/tex]:
- The coefficient of [tex]\(a^2\)[/tex] is 4.
- The coefficient of [tex]\(a\)[/tex] is 4.
- The constant term is -3.
We need to find pairs [tex]\((p, r)\)[/tex] such that:
1. The product of coefficients of [tex]\(a^2\)[/tex] and the constant term remain the same when multiplied: [tex]\(4 \times -3 = -12\)[/tex].
2. The middle term [tex]\(4a\)[/tex] can be broken down accordingly.
We need factors of -12 that add up to the middle coefficient 4. Possible pairs that multiply to -12 are:
- (1, -12)
- (-1, 12)
- (2, -6)
- (-2, 6)
- (3, -4)
- (-3, 4)
From these, we try:
[tex]\[
(ax + p)(qx + r) = 4a^2 + (ap + qr)a + 4a - 3
\][/tex]
Searching these pairs reveals:
Consider, -2, 6:
[tex]\[
(4a - 2)(a + 1.5) = 4a^2 + 6a – 3 (not correct)
\][/tex]
No clear set reveals apparent solution manually so:
Using factorization by grouping we identify:
First principle (AC method)
4a^2 + 4a - 3 = 4a^2 - 2a + 6a - 3 step proceed and break middle terms:
Factor out common terms:
= 2a (2a - 1) + 3 (2a - 1)
= (2a + 3) (2a - 1)
So, we conclude:
[tex]\[
4a^2 + 4a -3 = (4a+6)(2a)-1
when considered 2a for middle terms upto reveal in quadratic step which accurate till clear sqrt in pairing.
Thus most likely here :
\(4a^2 + 4a-3 = (2a - 1)(2a + 3)\)
So, the possible factorizations of \(4a^2 + 4a - 3\) include:
\[(2a - 1)(2a + 3)\][/tex]
