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Sure, let's break down the process of factoring the trinomial [tex]\( 12v^2 + 11v - 15 \)[/tex].
First, identify the coefficients in the trinomial [tex]\( 12v^2 + 11v - 15 \)[/tex]:
- The coefficient of [tex]\( v^2 \)[/tex] term (a) is 12.
- The coefficient of [tex]\( v \)[/tex] term (b) is 11.
- The constant term (c) is -15.
We want to factor this into the form [tex]\( (pv + q)(rv + s) \)[/tex]. To do this, we need to find numbers p, q, r, and s such that:
1. [tex]\( pr = 12 \)[/tex]
2. [tex]\( qs = -15 \)[/tex]
3. [tex]\( ps + qr = 11 \)[/tex]
From the calculations, we obtain the solution as follows:
1. The factors of 12 are 1, 2, 3, 4, and 6.
2. The factors of -15 are 1, -1, 3, -3, 5, and -5.
By trial and error (which might involve several steps if solving manually):
- We noticed that the correct pairs are [tex]\( p = 3 \)[/tex], [tex]\( q = 5 \)[/tex], [tex]\( r = 4 \)[/tex], and [tex]\( s = -3 \)[/tex].
Putting these together, the factored form of [tex]\( 12v^2 + 11v - 15 \)[/tex] is:
[tex]\[ (3v + 5)(4v - 3). \][/tex]
Therefore, the correct numbers to fill in the blanks are:
[tex]\[ (3v + 5)(4v - 3). \][/tex]
First, identify the coefficients in the trinomial [tex]\( 12v^2 + 11v - 15 \)[/tex]:
- The coefficient of [tex]\( v^2 \)[/tex] term (a) is 12.
- The coefficient of [tex]\( v \)[/tex] term (b) is 11.
- The constant term (c) is -15.
We want to factor this into the form [tex]\( (pv + q)(rv + s) \)[/tex]. To do this, we need to find numbers p, q, r, and s such that:
1. [tex]\( pr = 12 \)[/tex]
2. [tex]\( qs = -15 \)[/tex]
3. [tex]\( ps + qr = 11 \)[/tex]
From the calculations, we obtain the solution as follows:
1. The factors of 12 are 1, 2, 3, 4, and 6.
2. The factors of -15 are 1, -1, 3, -3, 5, and -5.
By trial and error (which might involve several steps if solving manually):
- We noticed that the correct pairs are [tex]\( p = 3 \)[/tex], [tex]\( q = 5 \)[/tex], [tex]\( r = 4 \)[/tex], and [tex]\( s = -3 \)[/tex].
Putting these together, the factored form of [tex]\( 12v^2 + 11v - 15 \)[/tex] is:
[tex]\[ (3v + 5)(4v - 3). \][/tex]
Therefore, the correct numbers to fill in the blanks are:
[tex]\[ (3v + 5)(4v - 3). \][/tex]
To factor the trinomial 12v + 11v - 15, we need to find two binomials (av + b) (cu + d) such that their product is the given trinomial.
1. First, consider the product of the leading coefficient (12) and the constant term (-15), which is -180.
2. We need to find two numbers that multiply to —180 and add up to the middle coefficient (11).
3. The numbers that satisfy these conditions are 20 and -9, since 20 × (-9) = -180
and 20 + (-9) = 11
4. Rewrite the middle term using these numbers:
1202 + 20v - 90 - 15
5. Factor by grouping:
• Group the first two terms and the last two terms:
(1202+200)+(-90-15).
• Factor out the greatest common factor from each group:
40(30 + 5) - 3(30 + 5)
6. Notice that (3v + 5) is a common factor:
(40=3)(30+5)
Thus, the factors of 1202 + 11v - 15 are (40 - 3) (30 + 5).
1. First, consider the product of the leading coefficient (12) and the constant term (-15), which is -180.
2. We need to find two numbers that multiply to —180 and add up to the middle coefficient (11).
3. The numbers that satisfy these conditions are 20 and -9, since 20 × (-9) = -180
and 20 + (-9) = 11
4. Rewrite the middle term using these numbers:
1202 + 20v - 90 - 15
5. Factor by grouping:
• Group the first two terms and the last two terms:
(1202+200)+(-90-15).
• Factor out the greatest common factor from each group:
40(30 + 5) - 3(30 + 5)
6. Notice that (3v + 5) is a common factor:
(40=3)(30+5)
Thus, the factors of 1202 + 11v - 15 are (40 - 3) (30 + 5).
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