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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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