can i have some quick help? this is in trinomial method
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#1
ac=6(-2)=-12
#2
#3
ac=-24
Answer:
(a) [tex](2x-3)(x+2)[/tex]
(b) [tex](5a-1)(2a+1)[/tex]
(c) [tex](2c-3)(c+4)[/tex]
Step-by-step explanation:
To factor a quadratic in the form [tex]ax^2+bx+c[/tex]
Part (a)
Given expression: [tex]2x^2+x-6[/tex]
[tex]\implies ac=2 \cdot -6=-12[/tex]
Factors of -12 that sum to 1: 4 and -3
[tex]\implies 2x^2+4x-3x-6[/tex]
Factor first two terms and last two terms separately:
[tex]\implies 2x(x+2)-3(x+2)[/tex]
Factor out common term [tex](x+2)[/tex]:
[tex]\implies (2x-3)(x+2)[/tex]
Part (b)
Given expression: [tex]10a^2+3a-1[/tex]
[tex]\implies ac=10 \cdot -1=-10[/tex]
Factors of -10 that sum to 3: 5 and -2
[tex]\implies 10a^2+5a-2a-1[/tex]
Factor first two terms and last two terms separately:
[tex]\implies 5a(2a+1)-1(2a+1)[/tex]
Factor out common term [tex](2a+1)[/tex]:
[tex]\implies (5a-1)(2a+1)[/tex]
Part (c)
Given expression: [tex]2c^2+5c-12[/tex]
[tex]\implies ac=2 \cdot -12=-24[/tex]
Factors of -24 that sum to 5: 8 and -3
[tex]\implies 2c^2+8c-3c-12[/tex]
Factor first two terms and last two terms separately:
[tex]\implies 2c(c+4)-3(c+4)[/tex]
Factor out common term [tex](c+4)[/tex]:
[tex]\implies (2c-3)(c+4)[/tex]