Answer :
Answer:
Sure, let's factor each of the given quadratic expressions:
Sure, let's factor each of the given quadratic expressions:
1. \(6x^2 + x - 2\):
First, we'll check if the expression can be factored using the quadratic formula:
\[x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\]
For \(6x^2 + x - 2\), \(a = 6\), \(b = 1\), and \(c = -2\).
Using the quadratic formula:
\[x = \frac{{-1 \pm \sqrt{{1^2 - 4(6)(-2)}}}}{{2(6)}}\]
\[x = \frac{{-1 \pm \sqrt{{1 + 48}}}}{{12}}\]
\[x = \frac{{-1 \pm \sqrt{49}}}{{12}}\]
\[x = \frac{{-1 \pm 7}}{{12}}\]
So, the solutions are \(x = \frac{6}{3} = 2\) and \(x = \frac{-8}{3} = -\frac{4}{3}\).
Now, we'll use these solutions to factor the quadratic expression:
\[6x^2 + x - 2 = 6(x - 2)(x + \frac{4}{3})\]
2. \(3x^2 - 14x - 24\):
First, we'll check if the expression can be factored using the quadratic formula:
\[x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\]
For \(3x^2 - 14x - 24\), \(a = 3\), \(b = -14\), and \(c = -24\).
Using the quadratic formula:
\[x = \frac{{14 \pm \sqrt{{(-14)^2 - 4(3)(-24)}}}}{{2(3)}}\]
\[x = \frac{{14 \pm \sqrt{{196 + 288}}}}{{6}}\]
\[x = \frac{{14 \pm \sqrt{484}}}{{6}}\]
\[x = \frac{{14 \pm 22}}{{6}}\]
So, the solutions are \(x = \frac{36}{6} = 6\) and \(x = \frac{-8}{3}\).
Now, we'll use these solutions to factor the quadratic expression:
\[3x^2 - 14x - 24 = 3(x - 6)(x + \frac{8}{3})\]
3. \(2x^2 + 9x + 10\):
First, we'll check if the expression can be factored using the quadratic formula:
\[x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\]
For \(2x^2 + 9x + 10\), \(a = 2\), \(b = 9\), and \(c = 10\).
Using the quadratic formula:
\[x = \frac{{-9 \pm \sqrt{{9^2 - 4(2)(10)}}}}{{2(2)}}\]
\[x = \frac{{-9 \pm \sqrt{{81 - 80}}}}{{4}}\]
\[x = \frac{{-9 \pm \sqrt{1}}}{{4}}\]
So, the solutions are \(x = \frac{-8}{4} = -2\) and \(x = \frac{-10}{4} = -\frac{5}{2}\).
Now, we'll use these solutions to factor the quadratic expression:
\[2x^2 + 9x + 10 = 2(x + 2)(x + \frac{5}{2})\]
1. \(6x^2 + x - 2\):
First, we'll check if the expression can be factored using the quadratic formula:
\[x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\]
For \(6x^2 + x - 2\), \(a = 6\), \(b = 1\), and \(c = -2\).
Using the quadratic formula:
\[x = \frac{{-1 \pm \sqrt{{1^2 - 4(6)(-2)}}}}{{2(6)}}\]
\[x = \frac{{-1 \pm \sqrt{{1 + 48}}}}{{12}}\]
\[x = \frac{{-1 \pm \sqrt{49}}}{{12}}\]
\[x = \frac{{-1 \pm 7}}{{12}}\]
So, the solutions are \(x = \frac{6}{3} = 2\) and \(x = \frac{-8}{3} = -\frac{4}{3}\).
Now, we'll use these solutions to factor the quadratic expression:
\[6x^2 + x - 2 = 6(x - 2)(x + \frac{4}{3})\]
2. \(3x^2 - 14x - 24\):
First, we'll check if the expression can be factored using the quadratic formula:
\[x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\]
For \(3x^2 - 14x - 24\), \(a = 3\), \(b = -14\), and \(c = -24\).
Using the quadratic formula:
\[x = \frac{{14 \pm \sqrt{{(-14)^2 - 4(3)(-24)}}}}{{2(3)}}\]
\[x = \frac{{14 \pm \sqrt{{196 + 288}}}}{{6}}\]
\[x = \frac{{14 \pm \sqrt{484}}}{{6}}\]
\[x = \frac{{14 \pm 22}}{{6}}\]
So, the solutions are \(x = \frac{36}{6} = 6\) and \(x = \frac{-8}{3}\).
Now, we'll use these solutions to factor the quadratic expression:
\[3x^2 - 14x - 24 = 3(x - 6)(x + \frac{8}{3})\]
3. \(2x^2 + 9x + 10\):
First, we'll check if the expression can be factored using the quadratic formula:
\[x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\]
For \(2x^2 + 9x + 10\), \(a = 2\), \(b = 9\), and \(c = 10\).
Using the quadratic formula:
\[x = \frac{{-9 \pm \sqrt{{9^2 - 4(2)(10)}}}}{{2(2)}}\]
\[x = \frac{{-9 \pm \sqrt{{81 - 80}}}}{{4}}\]
\[x = \frac{{-9 \pm \sqrt{1}}}{{4}}\]
So, the solutions are \(x = \frac{-8}{4} = -2\) and \(x = \frac{-10}{4} = -\frac{5}{2}\).
Now, we'll use these solutions to factor the quadratic expression:
\[2x^2 + 9x + 10 = 2(x + 2)(x + \frac{5}{2})\]
answers:
(2x-1)(3x+2)
(x-6)(3x+4)
(x+2)(2x+5)
i’ll attach a photo to better illustrate it without the ^s and such, but here is the explanation
first let’s establish the form ax^2+bx+c
first question: 6x^2+x-2
first we find ac (-12), then we need to find two numbers that multiply to -12 and add to b (1)
to find it you can list out multiples for -12
1, 2, 3, 4, 6, 12
4*-3 = -12; 4-3= 1 ✔️
then replace bx with the factors we found
6x^2 + 4x - 3x - 2
put parentheses (this works bc of commutative property)
(6x^2 +4x) (-3x-2)
factor these two pairs separately (you should get the same numbers left in the parentheses from each after factoring them)
2x ( 3x + 2 ) -1 ( 3x + 2 )
put the two factored out values into their own parentheses (AKA put in factored form)
(2x - 1)(3x+2) ✅
second question: 3x^2-14x-24
-24*3=-72
-72 = -18 * 4 & -18+4 = -14 ✔️
3x^2 - 18x + 4x - 24
(3x^2 - 18x) + (4x - 24)
3x(x-6) +4 (x-6)
(3x+4)(x-6) ✅
third question: 2x^2+9x+10
10*2=20
5*4 =20 & 5+4=9 ✔️
2x^2+5x+4x+10
(2x^2+5x)+(4x+10)
x(2x+5) 2(2x+5)
(x+2)(2x+5) ✅
hope this helps! let me know if you have any questions
(2x-1)(3x+2)
(x-6)(3x+4)
(x+2)(2x+5)
i’ll attach a photo to better illustrate it without the ^s and such, but here is the explanation
first let’s establish the form ax^2+bx+c
first question: 6x^2+x-2
first we find ac (-12), then we need to find two numbers that multiply to -12 and add to b (1)
to find it you can list out multiples for -12
1, 2, 3, 4, 6, 12
4*-3 = -12; 4-3= 1 ✔️
then replace bx with the factors we found
6x^2 + 4x - 3x - 2
put parentheses (this works bc of commutative property)
(6x^2 +4x) (-3x-2)
factor these two pairs separately (you should get the same numbers left in the parentheses from each after factoring them)
2x ( 3x + 2 ) -1 ( 3x + 2 )
put the two factored out values into their own parentheses (AKA put in factored form)
(2x - 1)(3x+2) ✅
second question: 3x^2-14x-24
-24*3=-72
-72 = -18 * 4 & -18+4 = -14 ✔️
3x^2 - 18x + 4x - 24
(3x^2 - 18x) + (4x - 24)
3x(x-6) +4 (x-6)
(3x+4)(x-6) ✅
third question: 2x^2+9x+10
10*2=20
5*4 =20 & 5+4=9 ✔️
2x^2+5x+4x+10
(2x^2+5x)+(4x+10)
x(2x+5) 2(2x+5)
(x+2)(2x+5) ✅
hope this helps! let me know if you have any questions
