4. Factorise :
(i) 12x2-7x+1
(iii
) 6x2 + 5x - 6
Answers
Answer:
HERE IS YOUR ANSWER:
(i)=12x^2-7x+1
\begin{gathered} = 12 {x}^{2} - 7x + 1 \\ = 12 {x}^{2} - 4x - 3x + 1 \\ = 4x(3x - 1) - 1(3x - 1) \\ = (3x - 1)(4x - 1) \\ \end{gathered}
=12x
2
−7x+1
=12x
2
−4x−3x+1
=4x(3x−1)−1(3x−1)
=(3x−1)(4x−1)
(ii)6x^2+5x-6
\begin{gathered} = 6 {x}^{2} + 5x - 6 \\ = 6 {x}^{2} + 9x - 4x - 6 \\ = 3x(2x + 3) - 2(2x + 3) \\ = (2x + 3)(3x - 2) \end{gathered}
=6x
2
+5x−6
=6x
2
+9x−4x−6
=3x(2x+3)−2(2x+3)
=(2x+3)(3x−2)
(iii)2x^2+7x+3
\begin{gathered} = 2 {x}^{2} + 7x + 3 \\ = 2 {x}^{2} + 6x + x + 3 \\ = 2x(x + 3)1( x+ 3) \\ = (x + 3)(2x + 1)\end{gathered}
=2x
2
+7x+3
=2x
2
+6x+x+3
=2x(x+3)1(x+3)
=(x+3)(2x+1)
(iv)3x^2-x-4
\begin{gathered} = 3 {x}^{2} - x - 4 \\ = 3 {x}^{2} - 4x + 3x - 4 \\ = x(3x - 4)1(3x - 4) \\ = (3x - 4)(x + 1)\end{gathered}
=3x
2
−x−4
=3x
2
−4x+3x−4
=x(3x−4)1(3x−4)
=(3x−4)(x+1)