Question

Division ( using factorise) Зm² + 22m + 35 by Зm + 7​

Answer 1

Step-by-step explanation:

$\huge\sf{solution}$

$3m {}^{2} + 22m + 35 \div 3m + 7$

$3 {m}^{2} + 15m + 7m + 35 \div 3m + 7$

$3m(m + 5) + 7(m + 5) \div 3m + 7$

$(3m + 7)(m + 5) \div 3m + 7$

$m + 5$

Answer 2

Answer:

In factorisation by simple division method, we first break the polynomial into its direct factors. For example, if we divide 8y3+7y2+6y by 2y, we first break the polynomial into its basic factors, i.e : 2y(4y) 2+ 2y (7/2 *y) + 2y(3)

Next, we write the common factor separately, where we get: 2y { 4y2+(7/2y) + 3}. In the last step, we divide the expression as asked in the question i.e: 2y {4y2+(7/2y) + 3} / 2y. The answer to this shall be: 4y2+ (7/2y) + 3

Example 1: Divide 16(x2yz + xy2z+xyz2) by 4xyz

Solution : 2×2×2×2× [(x×x×y×z) + (x×y×y×z) + (x×y×z×z)]

= 2×2×2×2× {x×y×z (x+y+z)}

= 16xyz (x+y+z)

Now divide the polynomial as given in the question:

= 4*4xyz (x+y+z) / 4xyz

= 4(x+y+z)