How do you factor the expression $25 y^{2} - 52 y + 27$ $25y^2- 52y + 27$ ?

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How do you factor the expression $25 y^{2} - 52 y + 27$ $25y^2- 52y + 27$ ?

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Answer 1 · 2018-05-21T05:17:23

$(25 y - 27) (y - 1)$ $(25y-27)(y-1)$

Explanation:

We need to find the factors that when multiplied it gives $675$ $675$ $(25 \times 27)$ $(25 xx 27)$ and when added it gives $- 52$ $-52$ .

Multiplying:
$675$ $675$ = $3 \times 3 \times 3 \times 5 \times 5$ $3xx3xx3xx5xx5$ = $27 \times 25$ $27 xx 25$

Adding:
$- 52$ $-52$ = $- 27 - 25$ $-27-25$
Hence, $- 27 \times - 25 = 675$ $-27 xx -25 = 675$

So the factors are: $- 27$ $-27$ and $- 25$ $-25$

$25 y^{2} - 52 y + 27$ $25y^2-52y+27$

$25 y^{2} - 25 y - 27 y + 27$ $25y^2-25y-27y+27$

$25 y (y - 1) - 27 (y - 1)$ $25y(y-1)-27(y-1)$

$(25 y - 27) (y - 1)$ $(25y-27)(y-1)$

Check the answer:
$(25 y - 27) (y - 1)$ $(25y-27)(y-1)$
$(25 y \times y) - 25 y - 27 y + 27$ $(25yxxy)-25y-27y+27$
$25 y^{2} - 52 y + 27$ $25y^2-52y+27$

Answer 2 · 2018-05-21T06:10:52

$25 y^{2} - 52 y + 27 = (y - 1) (25 y - 27)$ $25y^2-52y+27 = (y-1)(25y-27)$

Explanation:

Given:

$25 y^{2} - 52 y + 27$ $25y^2-52y+27$

Note that $25 - 52 + 27 = 0$ $25-52+27 = 0$

Hence $y = 1$ $y=1$ is a zero and $(y - 1)$ $(y-1)$ a factor.

The leading term of the other factor must be $25 y$ $25y$ to get $25 y^{2}$ $25y^2$ in the product and the trailing term must be $- 27$ $-27$ in order to get $+ 27$ $+27$ in the product.

So we find:

$25 y^{2} - 52 y + 27 = (y - 1) (25 y - 27)$ $25y^2-52y+27 = (y-1)(25y-27)$

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How do you factor the expression $25 y^{2} - 52 y + 27$ $25y^2- 52y + 27$ ?

2 Answers

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How do you factor the expression 25y^2- 52y + 2725y2−52y+2725y^2- 52y + 27?

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How do you factor the expression $25 y^{2} - 52 y + 27$ $25y^2- 52y + 27$ ?