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Ex.2.4 Q4 Polynomials Solution - NCERT Maths Class 9

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Question

Factorise:

(i)\(\begin{align}12{x^2} - 7x + 1 \end{align}\)

(ii)\(\begin{align}2{x^2} + 7x + 3 \end{align}\)

(iii) \(\begin{align}6{x^2} + 5x - 6 \end{align}\)

(iv) \(\begin{align}3{x^2} - x - 4 \end{align}\)

 Video Solution
Polynomials
Ex 2.4 | Question 4

Text Solution

Reasoning:

By splitting method, we can find factors using the following method.

Find \(2\) numbers \(p, q\) such that:

(i) \(p + q = \) co-efficient of \(x\)

(ii) \(pq = \) co-efficient of \({x^2}\) and the constant term.

Steps:

(i) \(\,12{x^2} - 7x + 1\)

\(p + q = - 7\)(co-efficient of \(x\))

\(pq = 12 \times 1 = 12\) (co-efficient of \({x^2}\,\) and the constant term.)

By trial and error method, we get \(p = -4, q = -3.\)

Now splitting the middle term of the given polynomial,

\[\begin{align}12{x^2} - 7x + 1 &= 12{x^2} - 4x - 3x + 1\\ &= 4x(3x - 1) - 1(3x - 1)\\ &= (3x - 1)(4x - 1) \\&\text {(taking} (3x\!-\!1\!)\!\text{ as common)}\end{align}\]

(ii) \(\,\,2{x^2} + 7x + 3\)

\(p + q = 7\) (co-efficient of \(x\))

\(pq = 2 \times 3 = 6\) (co-efficient of \({x^2}\,\) and the constant term.)

By trial and error method, we get \(p = 6, q = 1.\)

Now splitting the middle term of the given polynomial,

\[\begin{align}\,2{x^2} + 7x + 3 &= 2{x^2} + 6x + x + 3\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= 2x(x + 3) + 1(x + 3)\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= (2x + 1)(x + 3)\end{align}\]

(iii) \(6{x^2} + 5x - 6\)

\(p + q = 5\) (co-efficient of \(x\))

\(pq = 6 \times -6 = 36\) (co-efficient of \({x^2}\,\) and the constant term.)

By trial and error method, we get \(p = 9, q = -4.\)

Now splitting the middle term of the given polynomial,

\[\begin{align} 6{x^2} + 5x - 6 &= 6{x^2} + 9x - 4x - 6\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= 3x(2x + 3) - 2(2x + 3)\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= (3x - 2)(2x + 3)\end{align}\]

(iv) \(3{x^2} - x - 4\)

\(p + q = -1\) (co-efficient of \(x\))

\(pq = 3 \times -4 = -12\)(co-efficient of \({x^2}\,\) and the constant term.)

By trial and error method, we get \(p = -4, q = 3.\)

Now splitting the middle term of the given polynomial,

\[\begin{align} 3{x^2} - x - 4 &= 3{x^2} - 4x + 3x - 4\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= \,\,3{x^2} + 3x - 4x - 4\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= \,3x(x + 1) - 4(x + 1)\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= (3x - 4)(x + 1)\end{align}\]

 Video Solution
Polynomials
Ex 2.4 | Question 4
  
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