Here we are going to see some example problems on factoring quadratic equations.

 Pattern of quadratic equation Signs of factors ax2 + bx + cax2 - bx + cax2 - bx - cax2 + bx - c Both factors will have +Both factors will have -Greater number have - and smaller number have +Greater number have + and smaller number have -

Question 1 :

Factorise the following:

(i)  x2 + 10x + 24

Solution :

=  x2 + 10x + 24

Since the coefficient of x2 is 1, we have to take the constant 24. The given equation is in the form ax2 + bx + c, so both factors are positive.

24  =  6 ⋅ 4 and 6 + 4  =  10

=  x2 + 6x + 4x + 24

=  x(x + 6) + 4(x + 6)

=  (x + 6) (x + 4)

Hence the factors are (x + 6) and (x + 4).

(ii)  z2 + 4z - 12

Solution :

=  z2 + 4z - 12

Since the coefficient of x2 is 1, we have to take the constant -12. The given equation is in the form ax2 + bx - c, smaller factor will have negative.

-12  =  6 ⋅ (-2) and 6 + (-2)  =  4

=  x2 + 6x - 2x - 12

=  x(x + 6) - 2(x + 6)

=  (x - 2) (x + 6)

Hence the factors are (x - 2) and (x + 6)

Let us look into the next problem on "Factoring Quadratic Equation Examples".

(iii)  p2 - 6p - 16

Solution :

=  p2 - 6p - 16

Since the coefficient of x2 is 1, we have to take the constant -16. The given equation is in the form ax2 - bx - c, big number will have negative.

-16  =  -8 ⋅ 2 and -8 + 2  =  -6

=  x2 - 8x + 2x - 16

=  x(x - 8) + 2(x - 8)

=  (x + 2) (x - 8)

Hence the factors are (x + 2) and (x - 8)

(iv)  t2 + 72 - 17t

Solution :

=  t2 - 17p + 72

Since the coefficient of t2 is 1, we have to take the constant 72. The given equation is in the form ax2 - bx + c, both factors will have negative sign.

-72  =  -8 ⋅ (-9) and -8 + (-9)  =  -17

=  x2 - 8x - 9x + 72

=  x(x - 8) - 9(x - 8)

=  (x - 9) (x - 8)

Hence the factors are (x - 9) and (x - 8)

(v)  y2 - 16y - 80

Solution :

=  y2 - 16y - 80

Since the coefficient of y2 is 1, we have to take the constant -80. The given equation is in the form ax2 - bx - c, large factors will have negative sign.

-80  =  -20 ⋅ 4 and -20 + 4  =  -16

=  y2 - 20y + 4y - 80

=  y(y - 20) + 4(y - 20)

=  (y - 20) (y + 4)

Hence the factors are (y - 20) and (y + 4)

(vi)  a2 + 10a - 600

Solution :

=  a2 + 10a - 600

Since the coefficient of a2 is 1, we have to take the constant -600. The given equation is in the form ax2 + bx - c, small factors will have negative sign.

-600  =  30 ⋅ (-20) and 30 - 20  =  10

=  a2 + 30a - 20a - 600

=  a(a + 30) - 20(a + 30)

=  (a - 20)(a + 30)

Hence the factors are (a - 20) and (a + 30).

After having gone through the stuff given above, we hope that the students would have understood, "Factoring Quadratic Equation Examples"

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